Feynman checkerboard
An intro to algorithmic quantum field theory
E. Akhmedova, M. Skopenkov, A. Ustinov, R. Valieva, A. Voropaev
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Basic model from Β§1
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Model with mass from Β§3
Figure 1: The probability tofind an electron at a given point(white depicts large oscillations)
Summary. It turns out that rainbow patterns on soapbubbles andunbelievable laws of motion of electrons can be explained by means ofone simple-minded model. It is a game, in which a checker moves on acheckerboard by certain simple rules, and we count the turnings. ThisβFeynman checkerboardβ can explain all phenomena in the world (witha serious proviso) except atomic nuclei and gravitation. We are goingto solve mathematical problems related to the game and discuss theirphysical meaning; no knowledge of physics is assumed.
Main results. The main results are explicit formulae for
β the percentage of light of given color reflected by a glass plate ofgiven width (Problem 24);
β the probability to find an electron at a given point, if it was emit-ted from the origin (Problem 15; see Figure 1).
Although the results are stated in physical terms, they are mathemati-cal theorems, because we provide mathematical models of the physicalphenomena in question, with all the objects defined rigorously. Moreprecisely, a sequence of models with increasing precision.
Plan. We start with a basic model and upgrade it step by stepin each subsequent section. Before each upgrade, we summarize whichphysical question does it address, which simplifying assumptions doesit resolve or impose additionally, and which experimental results doesit explain. Our aim is what is called 2-dimensional quantum electro-dynamics but the last steps on this way (sketched in Sections 8β10)still have not been done. (A 4-dimensional one can already explain allphenomena β with proviso and exceptions β but we do not discuss it.)
The scheme of upgrades dependence might help to choose your way:
1. Basic model
))
6. External field // 8***. Interaction
**
4. Source
οΏ½οΏ½
2. Spin //oo
οΏ½οΏ½
OO
7. Identical particles & antiparticles
οΏ½οΏ½
OO
10***. QED
5. Medium 3. Mass //oo 9***. Creation & annihilation
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Conventions. If a problem is a statement of an assertion, then it is requested to prove the assertion.A puzzle is a problem, in which both a precise statement and a proof are requested. Hard problems aremarked with stars; you get first rank in Feynman checkerboard for solving one, and masters in Feynmancheckerboard for solving three from three different sections. Solutions are accepted in writing but you mayspend an earned star for 10 attempts to submit a solution in oral form (successful or not). If you cannotsolve a problem, proceed to the next ones: they may provide hints. Even if you do not reach the top(the proofs of main results), you learn much. You are encouraged to state and try to prove also your ownobservations and conjectures; you become a grand-master in Feynman checkerboard for discovering a newnontrivial one (and maybe even write your own scientific paper).
1
1 Basic modelQuestion: what is the probability to find an electron at the point (π₯, π¦), if it was emitted from the point (0, 0)?Assumptions: no self-interaction, no creation of electron-positron pairs, unit mass and lattice step, point source;no nuclear forces, no gravitation, electron moves uniformly along the π¦-axis and does not move along the π§-axis.Results: double-slit experiment, charge conservation.
Figure 2: Checker paths
On an infinite checkerboard, a checker moves to the diagonal-neighboring squares, either upwards-right or upwards-left. To each pathπ of the checker, assign a vector οΏ½οΏ½(π ) as follows. Start with a vector oflength 1 directed upwards. While the checker moves straightly, the vectoris not changed, but each time when the checker changes the direction, thevector is rotated through 90β clockwise (independently of the direction thechecker turns). In addition, at the very end the vector is divided by 2(π¦β1)/2,where π¦ is the total number of moves. The final position of the vector iswhat we denote by οΏ½οΏ½(π ). For instance, for the path in Figure 2 to the top,the vector οΏ½οΏ½(π ) = (1/8, 0) is directed to the right and has length 1/8.
Denote οΏ½οΏ½(π₯, π¦) :=β
π οΏ½οΏ½(π ), where the sum is over all the paths of thechecker from the square (0, 0) to the square (π₯, π¦), starting with the upwards-right move. Set οΏ½οΏ½(π₯, π¦) := 0, if there are no such paths. For instance,οΏ½οΏ½(1, 3) = (0,β1/2)+(1/2, 0) = (1/2,β1/2). The length square of the vectorοΏ½οΏ½(π₯, π¦) is called the probability1 to find an electron in the square (π₯, π¦), if itwas emitted from the square (0, 0). Notation: π (π₯, π¦) := |π(π₯, π¦)|2.
In Figure 1 to the top, the color of a point (π₯, π¦) with even π₯+π¦ depictsthe value π (π₯, π¦). The sides of the apparent angle are not the lines π¦ = Β±π₯ (and nobody knows why!).
In what follows squares (π₯, π¦) with even and odd π₯ + π¦ are called black and white respectively.
1. Observations for small π¦. Answer the following questions for each π¦ = 1, 2, 3, 4 (and state your ownquestions and conjectures for arbitrary π¦): Find the vector οΏ½οΏ½(π₯, π¦) and the probability π (π₯, π¦) for each π₯.When π (π₯, π¦) = 0? What is
βπ₯βZ π (π₯, π¦) for fixed π¦? What are the directions of οΏ½οΏ½(1, π¦) and οΏ½οΏ½(0, π¦)?
The probability2 to find an electron in the square (π₯, π¦) subject to absorption in the square (π₯β², π¦β²) isdefined analogously to π (π₯, π¦), only the summation is over those paths π that do not pass through (π₯β², π¦β²).The probability is denoted by π (π₯, π¦ bypass π₯β², π¦β²).
-1000 -500 500 1000
-0.05
0.05
Figure 3: π2(π₯, 1000)2. Double-slit experiment. Is it true that π (π₯, π¦) = π (π₯, π¦ bypass 0, 2)+π (π₯, π¦ bypass 2, 2)? Is it true that π (π₯, π¦) β₯ π (π₯, π¦ bypass π₯β², π¦β²)?
3. Find π (0, 12). How to table the values οΏ½οΏ½(π₯, π¦) quickly without exhaustionof all paths? (The first solution grants first rank in Feynman checkers.)
Denote by π1(π₯, π¦) and π2(π₯, π¦) the coordinates of οΏ½οΏ½(π₯, π¦); see Figure 3.
4. Diracβs equation. Express π1(π₯, π¦) and π2(π₯, π¦) through π1(π₯Β± 1, π¦ β 1) and π2(π₯Β± 1, π¦ β 1).
5. Probability/charge conservation. For each positive integer π¦ we haveβ
π₯βZ π (π₯, π¦) = 1.
6. Symmetry. How the values π1(π₯, 100) for π₯ < 0 and for π₯ β₯ 0 are related with each other? The samefor the values π1(π₯, 100) + π2(π₯, 100).
7. Huygensβ principle. What is a fast way to find οΏ½οΏ½(π₯, 199), if we know οΏ½οΏ½(π₯, 100) for all integers π₯?
8. Using a computer, plot the graphs of the functions ππ¦(π₯) = π (π₯, π¦) for various π¦, joining each pair ofpoints (π₯, ππ¦(π₯)) and (π₯ + 2, ππ¦(π₯ + 2)) by a segment; cf. Figure 3. The same for the function π1(π₯, π¦).
9.* Find an explicit formula for the vector οΏ½οΏ½(π₯, π¦) and the probability π (π₯, π¦) (it is allowed to use a sumwith at most π¦ summands in the answer).
10.** (Skipable puzzle) Guess a simple βapproximate formulaβ for οΏ½οΏ½(π₯, π¦) and π (π₯, π¦), accurate for |π₯| βͺ π¦.1One should think of the value π¦ as fixed, and the squares (βπ¦, π¦), (βπ¦ + 2, π¦), . . . , (π¦, π¦) as all the possible outcomes of
an experiment. For instance, the π¦-th horizontal might be a photoplate detecting the electron.Familiarity with probability theory is not required for solving the presented problems.Beware that our rule for the probability computation is valid only for the basic model in question; we are going to changethe rule slightly in the upgrades. We make similar remarks each time we break some fundamental principles for simplicity.
2Thus an additional outcome of the experiment is that the electron has been absorbed and has not reached the photoplate.
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2 Spin
Question: what is the probability to find a right electron at (π₯, π¦), if a right electron was emitted from (0, 0)?Assumptions: the same.Results: spin reversal.
The trick in the solution of the previous problems has a physical meaning: it is convenient to consideran electron as being in one of the two states: right-moving or left-moving. We write just βright β or βleft βfor brevity3. This is not just a convenience but reflects an inalienable electronβs property called spin4.
Denote οΏ½οΏ½(π₯, π¦,+) :=β
π οΏ½οΏ½(π ), where the sum is over only those paths from (0, 0) to (π₯, π¦), which bothstart and finish with an upwards-right move. Define οΏ½οΏ½(π₯, π¦,β) to be an analogous sum over paths whichstart with an upwards-right move but finish with an upwards-left move.
The length square of the vector οΏ½οΏ½(π₯, π¦,+) (respectively, οΏ½οΏ½(π₯, π¦,β)) is called the probability5 to find aright (respectively, left) electron in the square (π₯, π¦), if a right electron was emitted from the square (0, 0).Denote by π (π₯, π¦,+) := |π(π₯, π¦,+)|2 and π (π₯, π¦,β) := |π(π₯, π¦,β)|2 these probabilities.
11. Express οΏ½οΏ½(π₯, π¦,+) and οΏ½οΏ½(π₯, π¦,β) through π1(π₯, π¦) and π2(π₯, π¦); π (π₯, π¦) through π (π₯, π¦,+) and π (π₯, π¦,β).
12.* Spin reversal. What is the probability π (π¦0,β) :=β
π₯βZ π (π₯, π¦0,β) to find a left electron on theline π¦ = π¦0 (it is allowed to use a sum with at most π¦0 summands in the answer)? Find the maximalelement and the limit of the sequence π (1,β), π (2,β), π (3,β), . . . .
A crash-course in calculus. For this and other problems with stars, the following formuli might be useful.Recall that
βππ=0
12π
= 2β 12π . Clearly, as π increases, the sum becomes closer and closer to 2. We would like
to writeββ
π=012π
= 2; let us give a definition of such an infinite sum. A sequence π1, π2, π3, . . . has a limit π, iffor each real π > 0 there is π such that for each integer π > π we have |ππ β π| < π. Notation: π = limπββ ππ.For instance, limπββ
(2β 1
2π
)= 2. By definition, put
ββπ=0 ππ = limπββ
βππ=0 ππ. Then indeed
ββπ=0
12π
= 2.The following generalization is called Newtonβs binomial theorem (allowed to use without proof):
(1 + π₯)π =
ββπ=0
π(π β 1) Β· Β· Β· (π β π + 1)
π(π β 1) Β· Β· Β· 1π₯π
for each complex π₯ with |π₯| < 1 and each real π, or for π₯ = 1 and π > β1. In particular, for π = β1 and β12 we get
1
1β π₯=
ββπ=0
π₯π and1β1β π₯
=ββπ=0
2π(2π β 1) Β· Β· Β· (π + 1)
π(π β 1) Β· Β· Β· 1π₯π
4π.
The following Stirling formula allows to estimate the summands in Newtonβs binomial theorem:
β2π ππ+1/2πβπ β€ π(π β 1) Β· Β· Β· 1 β€ π ππ+1/2πβπ.
Here π denotes limπββ(1 + 1/π)π. It is an irrational number between 2.71 and 2.72.
3 MassQuestion: what is the probability to find a right electron of mass π at (π₯, π¦), if it was emitted from (0, 0)?Assumptions: the mass and the lattice step are now arbitrary.Results: a formula for the probability for small lattice step.
To check our model against experiment we need the following generalization.Fix π,π > 0 called lattice step and particle mass respectively. To each path π of the checker, assign
a vector οΏ½οΏ½(π ,ππ) as follows. Start with the vector (0, 1). While the checker moves straightly, the vectoris not changed, but each time when the checker changes the direction, the vector is rotated through 90β
3Beware that in 3 or more dimensions βright β and βleft β mean something very different from the movement direction.Although often visualized as the direction of the electron rotation, these states cannot be explained in nonquantum terms.
4And chirality ; beware that the term spin usually refers to a property, not related to the movement direction at all.5Thus an experiment outcome is a pair (final π₯-coordinate, last-move direction), whereas the final π¦-coordinate is fixed.
These are the fundamental probabilities, whereas π (π₯, π¦) should in general be defined by the formula from the solution ofProblem 11 rather than the above formula π (π₯, π¦) = |π(π₯, π¦)|2 (being a coincidence).
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clockwise and multiplied by ππ. In addition, at the very end the vector is divided by (1 + π2π2)(π¦β1)/2,where π¦ is the total number of moves. The final position of the vector is what we denote by οΏ½οΏ½(π ,ππ). Thevectors οΏ½οΏ½(π₯, π¦,ππ,Β±) and the numbers π (π₯, π¦,ππ,Β±) are defined analogously to οΏ½οΏ½(π₯, π¦,Β±) and π (π₯, π¦,Β±),only οΏ½οΏ½(π ) is replaced by οΏ½οΏ½(π ,ππ). For instance, π (π₯, π¦, 1,+) = π (π₯, π¦,+). In Figure 1 to the bottom, thecolor of a point (π₯, π¦) with even π₯ + π¦ depicts the value π (π₯, π¦, 0.02,+) + π (π₯, π¦, 0.02,β).
13. (Puzzle) Massless and heavy particles. Find π (π₯, π¦, 0,+) and define π (π₯, π¦,β,+) for each π₯, π¦.
14. Solve analogues of Problems 4, 5, and 9 for ππ = 1.
In an experiment, we measure the probabilities to find the electron in intervals π₯0 β€ π₯ β€ π₯0 + βπ₯,π¦ = π¦0 rather than at particular points. Here π₯0, π¦0,βπ₯ are not integers but actual lengths measured inmeters. If all squares have small size 1
πΓ 1
π, then the interval is approximated by the collection of black
squares(2βππ₯2
β, 2
βππ¦2
β)with π₯, π¦ satisfying the above (in)equalities. This leads to the following problem.6
15.* (First main problem) Continuum limit. For each π₯, π¦,π find limπββ π οΏ½οΏ½(2βππ₯2
β, 2βππ¦2
β, ππ,β
)and
limπββ π οΏ½οΏ½(2βππ₯2
β, 2βππ¦2
β, ππ,+
)In the answer, it is allowed to use the following expressions7
π½0(π§) :=ββπ=0
(β1)π(π§/2)2π
(π!)2and π½1(π§) :=
ββπ=0
(β1)π(π§/2)2π+1
π!(π + 1)!.
4 SourceQuestion: what is the probability to find a right electron at (π₯, π¦), if it was emitted by a source of wave length π?Assumptions: the source is now realistic.Results: wave propagation.
Figure 4: Checker paths maynow start at distinct squares
A realistic source does not produce electrons localized at π₯ = 0 (asin our game) but a rather wide wave impulse instead. For our game,this means that the checker can start from an arbitrary black square onthe horizontal line π¦ = 0 (not too far from the origin), but the initialdirection of the vector οΏ½οΏ½(π ) is rotated through an angle proportional tothe distance from the starting square to the origin; see Figure 4.
Formally, fix real π, π > 0 and odd β called lattice step, wave length,and impulse width respectively. Denote by π πΌ οΏ½οΏ½ the rotation of a vector οΏ½οΏ½through the angle |πΌ|, which is counterclockwise for πΌ β₯ 0 and clockwisefor πΌ < 0. Define the vector
οΏ½οΏ½(π₯, π¦, π/π,β,+) :=1ββ
Ξβ1βπ₯0=1βΞπ₯0 even
βπ
π 2ππ₯0π/π οΏ½οΏ½(π ),
where the second sum is over all checker paths π from the square (π₯0, 0) to the square (π₯, π¦), starting andending with an upwards-right move. The length square of the vector is the probability to find a right electronat (π₯, π¦), emitted by a source of wave length π and impulse width β. It is denoted by π (π₯, π¦, π/π,β,+).Define οΏ½οΏ½(π₯, π¦, π/π,β,β) and π (π₯, π¦, π/π,β,β) analogously. For instance, π (π₯, π¦, π/π, 1,+) = π (π₯, π¦,+)for each π, π, and οΏ½οΏ½(π₯ + 1, 1, π/π,β,+) = 1β
Ξ
(β sin 2ππ₯π
π, cos 2ππ₯π
π
)for even |π₯| < β.
16. Let β = 3, π/π = 4. Find the vector οΏ½οΏ½(π₯, π¦, 4, 3,+) and the probability π (π₯, π¦, 4, 3,+) for π¦ = 1, 2, 3and each π₯. What is
βπ₯βZ(π (π₯, π¦, 4, 3,+)+π (π₯, π¦, 4, 3,β)) for fixed π¦ = 1, 2, 3? When π (π₯, 3, 4, 3,+) = 0?
17. Probability/charge conservation. Solve analogues of Problems 4 and 5 for οΏ½οΏ½(π₯, π¦, π/π,β,β),οΏ½οΏ½(π₯, π¦, π/π,β,+), and π (π₯, π¦, π/π,β,+) + π (π₯, π¦, π/π,β,β) instead of π1(π₯, π¦), π2(π₯, π¦), and π (π₯, π¦).
18. Causality. Bothβ
βοΏ½οΏ½(π₯, π¦, π/π,β,+) and βπ (π₯, π¦, π/π,β,+) do not depend on β for β > π¦ + |π₯|.Denote8 οΏ½οΏ½(π₯, π¦, π/π,Β±)=
ββοΏ½οΏ½(π₯, π¦, π/π,β,Β±) and π (π₯, π¦, π/π,Β±)=βπ (π₯, π¦, π/π,β,Β±) for any β>π¦+|π₯|.
19. Wave. How to find οΏ½οΏ½(π₯, 100, π/π,+) for all even π₯, if we know it for just one even π₯?
20.* Wave propagation. For each π₯, π¦, π, π find οΏ½οΏ½(π₯, π¦, π/π,β), π (π₯, π¦, π/π,β), π (π₯, π¦, π/π,β)+π (π₯, π¦, π/π,+).6In the limits, the normalization factor of π is a bit harder to explain; we not discuss it.7Called Bessel functions, which are almost as well-studied as sine and cosine; but familiarity with them is not required.8The notation οΏ½οΏ½(π₯, π¦, π/π,Β±) should not be confused with οΏ½οΏ½(π₯, π¦,ππ,Β±).
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5 MediumQuestion: which part of light of given color reflects from a glass plate of given width?Assumptions: right angle of incidence, no polarization of light; the mass now depends on π₯ but not on the color.Results: thin-film reflection.
Our model can be applied also to describe propagation of light in transparent media such as glass9.Light propagates as if it had some nonzero mass inside the media, while the mass remains zero outside10.The wavelength determines the color of light.
21. (Puzzle) Define an analogue of οΏ½οΏ½(π ,ππ) in the case when the mass π = π(π₯) depends on π₯ so thatanalogues of Problems 4 and 5 remain true.
Given a mass π = π(π₯), define οΏ½οΏ½(π₯, π¦,π(π₯)π, π/π,Β±) and π (π₯, π¦,π(π₯)π, π/π,Β±) analogously toοΏ½οΏ½(π₯, π¦, π/π,Β±) and π (π₯, π¦, π/π,Β±), only replace οΏ½οΏ½(π ) by οΏ½οΏ½(π ,ππ) in the definition. First we take π = 1.
22. One-surface reflection. Find π (π₯, π¦,π(π₯), π,+) and π (π₯, π¦,π(π₯), π,β) for
π(π₯) = π0(π₯) β‘ 0 and π(π₯) = π1(π₯) =
{0.2, for π₯ = 0;
0, for π₯ = 0.
Now fix odd πΏ > 1 called the width of a glass plate. First assume for simplicity that light is reflectedonly by the two surfaces of the plate and thus take11
π2(π₯)=
β§βͺβ¨βͺβ©β0.2, for π₯ = 1;
+0.2, for π₯ = πΏ;
0, otherwise.
The reflection/transmission probabilities12 of light of wavelength π by glass plate of width πΏ are respectivelyπ (π, πΏ,β) = lim
π¦β+βπ¦ even
π (0, π¦,π2(π₯), π,β);
π (π, πΏ,+) = limπ¦β+βπ¦ even
π (πΏ + 1, π¦,π2(π₯), π,+).
23. Plot the graph of the function π(πΏ) = π (0, 2πΏ,π2(π₯), 16,β).
24.* (Second main problem)Two-surface reflection. Find π (π, πΏ,β), π (π, πΏ,+), π (π, πΏ,β)+π (π, πΏ,+),and maxπΏ π (π, πΏ,β).
In fact the light is reflected inside the plate; this should be taken into account for a more accuratecomputation of the reflection probability (there is no hope for an exact solution for complicated matter suchas glass). For this purpose we need to modify the model essentially. Take arbitrary π > 0. Fix π > 0 andlet π3(π₯) = π for 0 < π₯ β€ πΏ/π and π3(π₯) = 0 otherwise. Now for each move starting in a square insidethe glass, our vector is additionally rotated through the angle arctanππ clockwise (independently on if thechecker does or does not turn in the square)13. In other words, set οΏ½οΏ½m(π ,π(π₯)π) := π βπ arctanπποΏ½οΏ½(π ,π(π₯)π),where π is the number of moves starting in the strip 0 < π₯ β€ πΏ/π in the path π . Define πm(π₯, π¦,π(π₯)π, π
π,β)
analogously to π (π₯, π¦,π(π₯)π, ππ,β), only replace οΏ½οΏ½(π ,π(π₯)π) by οΏ½οΏ½m(π ,π(π₯)π) in the definition.
25.** Thin-film reflection. Find limπβ0+
limπ¦β+βπ¦ even
πm(0,π¦,π3(π₯)π,ππ,β). For which π maximum of the expres-
sion over πΏ equals maxπΏ π (π, πΏ,β)? (Use existence of limπ¦β+βπ¦+π₯ even
π 2ππ¦π/ποΏ½οΏ½(π₯,π¦,π3(π₯)π,ππ,β) without proof.)
9Beware: in general Feynman checkerboard is inappropriate to describe light; partial reflection is a remarkable exception.10The mass is proportional to (πβ 1)/2
βπ, where π is the refractive index; for glass π β 1.5 and (πβ 1)/2
βπ β 0.2.
11This simplifying assumption requires negative mass for the left surface; the origin of that becomes clear after solving 25.12It is more conceptual to define π (π, πΏ,β) = limπβ0 limΞβ+β limπ¦β+β
βπ₯βZ π (π₯, π¦,π3(π₯)π,
ππ ,Ξ,β) but we do not.
13This additional rotation is explained as follows. The light can be scattered in each square inside the glass several times.Each individual scattering gives a factor of βπππ to our vector (viewed as a complex number) and may or may not changethe movement direction. Assume that ππ < 1. Thus a move without changing the direction contributes a factor of
1 (no scattering)β πππ (1 scattering)+ (βπππ)2 (2 scatterings)+ Β· Β· Β· = 1
1 + πππ.
Without a scattering, the checker moves straightly. Thus a turn in a particular square inside the glass contributes a factor
βπππ (1 scattering)+ (βπππ)2 (2 scatterings)+ (βπππ)3 (3 scatterings)+ Β· Β· Β· = βπππ
1 + πππ.
These are the same factors as in the model from Β§3 but additionally rotated through the angle arctanππ clockwise.
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6 External fieldQuestion: what is the probability to find a right electron at (π₯, π¦), if it moves in a given magnetic field π’?Assumptions: the magnetic field vanishes outside the π₯π¦-plane, it is not affected by the electron.Results: deflection of electron and spin βprecessionβ in a magnetic field, charge conservation.
Figure 5: Paths in a field
An external magnetic field changes the motion as follows14.A common point of 4 squares of the checkerboard is called a ver-
tex. A magnetic field15 is a fixed assignment π’ of numbers +1 andβ1 to all the vertices. For instance, in Figure 5, the magnetic fieldis β1 at the top-right vertex of each square (π₯, π¦) with both π₯ andπ¦ even. Modify the definition of the vector οΏ½οΏ½(π ) by reversing the di-rection each time when the checker passes through a vertex with themagnetic field β1. Denote by οΏ½οΏ½(π , π’) the resulting vector. Formally,put οΏ½οΏ½(π , π’) = οΏ½οΏ½(π )π’(πΆ1)π’(πΆ2) . . . π’(πΆπ¦), where πΆ1, πΆ2, . . . , πΆπ¦ are allthe vertices passed by π . Define οΏ½οΏ½(π₯, π¦, π’,Β±) and π (π₯, π¦, π’,Β±) analogously to οΏ½οΏ½(π₯, π¦,Β±) and π (π₯, π¦,Β±) re-placing οΏ½οΏ½(π ) by οΏ½οΏ½(π , π’) in the definition. For instance, if π’(πΆ) = +1 identically, then π (π₯, π¦, π’) = π (π₯, π¦).
26. Homogeneous field. Let π’(πΆ) = β1, if πΆ is the top-right vertex of a square (π₯, π¦) with both π₯and π¦ even, and π’(πΆ) = +1 otherwise. Find the vector οΏ½οΏ½(π₯, π¦, π’,+) and the probability π (π₯, π¦, π’,+) forπ¦ = 1, 2, 3, 4 and each integer π₯. What is
βπ₯βZ(π (π₯, π¦, π’,+) + π (π₯, π¦, π’,β)) for fixed π¦ = 1, 2, 3, or 4?
27. Spin βprecessionβ in a magnetic field. Plot the graph of the function π(π¦) =β
π₯βZ π (π₯, π¦, π’,+)for the field π’ from the previous problem using a computer.
For a given magnetic field π’, a white square is negative, if π’ equals β1 at 1 or 3 vertices of the square.
28. Gauge transformations. Changing the signs of the values of π’ at the 4 vertices of one black squaresimultaneously does not change π (π₯, π¦, π’,+).
29. Curvature. One can make π’ to be identically +1 in a rectangle formed by checkerboard squares usingthe transformations from Problem 28, if and only if there are no negative white squares in the rectangle.
30. Homology. The magnetic field π’ equals +1 on the boundary of a rectangle π Γ π formed bycheckerboard squares. Which can be the number of negative white squares in the rectangle?
31. Probability/charge conservation. Solve analogues of Problems 4, 5 for π’ not being identically +1.
7 Identical particles and antiparticles
Question: what is the probability to find electrons (or electron+positron) at πΉ and πΉ β², emitted from π΄ and π΄β²?Assumptions: the same as in the basic model; π¦-coordinate is interpreted as time.Results: exclusion principle.
The motion of several electrons is described by a similar model as follows.To each pair of paths π , π β² of the checker, consisting of π¦ moves each, assign a vector οΏ½οΏ½(π , π β²) as follows.
Start with the vector (0, 1). Move the checker consecutively along both paths, and rotate the vectoraccording to the same rule as in Β§1: each time when the checker changes the direction, the vector isrotated through 90β clockwise. (Thus the vector is rotated totally π‘(π ) + π‘(π β²) times, where π‘(π) is thenumber of turns in a path π.) In addition, at the very end the vector is divided by 2π¦β1. The final positionof the vector is denoted by οΏ½οΏ½(π , π β²). For instance, in Figure 2 we have οΏ½οΏ½(π , π 0) = (β1/4, 0).
Fix squares π΄ = (0, 0), π΄β² = (π₯0, 0), πΉ = (π₯, π¦), πΉ β² = (π₯β², π¦) and their diagonal neighbors π΅ = (1, 1),π΅β² = (π₯0 + 1, 1), πΈ = (π₯β 1, π¦ β 1), πΈ β² = (π₯β² β 1, π¦ β 1), where π₯0 = 0. Denote16
οΏ½οΏ½(π΄π΅,π΄β²π΅β² β πΈπΉ,πΈ β²πΉ β²) :=β
π :π΄π΅βπΈπΉπ β²:π΄β²π΅β²βπΈβ²πΉ β²
οΏ½οΏ½(π , π β²) ββ
π :π΄π΅βπΈβ²πΉ β²π β²:π΄β²π΅β²βπΈπΉ
οΏ½οΏ½(π , π β²),
14Beware that this method of adding the magnetic field, although well-known, is very different from the one from [Feynman].15Or electromagnetic vector-potential, to be precise. The field is interpreted as magnetic or electric depending on if the
π¦-coordinate is interpreted as position or time.16Here it is essential that π and π β² are paths of particles of the same sort, e.g., two electrons. Otherwise the 2nd sum is
omitted. The sign before the 2nd sum is changed to plus for some other sorts of particles, e.g., photons (particles of light).
6
where the first sum is over all pairs consisting of a checker path π starting with the move π΄π΅ and endingwith the move πΈπΉ , and a path π β² starting with the move π΄β²π΅β² and ending with the move πΈ β²πΉ β², whereasin the second sum the final moves are interchanged.
The length square π (π΄π΅,π΄β²π΅β² β πΈπΉ,πΈ β²πΉ β²) = |π(π΄π΅,π΄β²π΅β² β πΈπΉ,πΈ β²πΉ β²)|2 is called the probability17
to find right electrons at πΉ and πΉ β², if they are emitted from π΄ and π΄β². In particular, π (π΄π΅,π΄β²π΅β² βπΈπΉ,πΈπΉ ) = 0, i.e., two right electrons cannot be found at the same point; this is called exclusion principle.
32. Independence. For π₯0 β₯ 2π¦ and π₯β² > π₯ express π (π΄π΅,π΄β²π΅β² β πΈπΉ,πΈ β²πΉ β²) through π (π₯, π¦,+) andπ (π₯β² β π₯0, π¦,+).
33. Exclusion principle (for intermediate states). Prove that οΏ½οΏ½(π΄π΅,π΄β²π΅β² β πΈπΉ,πΈ β²πΉ β²) is notchanged, if the sums in the definition are over only those pairs of paths π , π β² which have no common moves.
Define π (π΄π΅,π΄β²π΅β² β πΈπΉ,πΈ β²πΉ β²) analogously also for πΈ = (π₯Β± 1, π¦ β 1), πΈ β² = (π₯β² Β± 1, π¦ β 1).
34. Probability/charge conservation. We haveβ
πΈ,πΈβ²,πΉ,πΉ β² π (π΄π΅,π΄β²π΅β² β πΈπΉ,πΈ β²πΉ β²) = 1, where thesum is over all quadruples πΉ = (π₯, π¦), πΉ β² = (π₯β², π¦), πΈ = (π₯Β±1, π¦β1), πΈ β² = (π₯β²Β±1, π¦β1) with fixed π¦ β₯ 1.
An electron has an antiparticle called positron. One can think of an antiparticle as a particle movingbackwards in time. The motion is described by a similar model as follows.
Add one more checker to the checkerboard, which now moves either downwards-right or downwards-leftto the diagonal neighbors. The added checker is called black, while the one studied before is called white.
For each pair of paths π , π β² of the white and black checker respectively, define a vector οΏ½οΏ½(π , π β²) analogouslyto the above, only for each turn of the black checker rotate the vector counterclockwise rather than clockwise(independently of the direction the checker turns). For instance, οΏ½οΏ½(π , π β²) = (0, 21βπ¦), if π and π β² is the samepath with opposite directions. Denote18 by
οΏ½οΏ½(π΄π΅,π΅β²π΄β² β πΈπΉ, πΉ β²πΈ β²) :=βπ ,π β²
οΏ½οΏ½(π , π β²)
the sum is over all the pairs of a white-checker path π starting with the move π΄π΅ and ending with themove πΈπΉ , and a black-checker path π β² starting with the move πΉ β²πΈ β² and ending with the move π΅β²π΄β². Thelength square π (π΄π΅,π΅β²π΄β² β πΈπΉ, πΉ β²πΈ β²) = |π(π΄π΅,π΅β²π΄β² β πΈπΉ, πΉ β²πΈ β²)|2 is called the probability to find anelectron at πΉ and a positron at πΉ β², if they are emitted from π΄ and π΄β².
35. Independence. Solve analogue of Problem 32 for π (π΄π΅,π΅β²π΄β² β πΈπΉ, πΉ β²πΈ β²).
8 Interaction***Question: the same as in the basic model but the π¦-coordinate is now interpreted as time.Assumptions: the electron now generates a magnetic field affecting the motion; encircling by waveproof walls.Results: interaction with the walls changes the motion of an electron.
We have reached an unexplored area: a physically correct rigorous definition of the next upgrade isnot yet known. In this section we give a WRONG definition leading to paradoxical results:
β the upgrades measures the interaction of the electron with the walls rather than self-interaction;
β the interaction (with the walls) propagates at infinite speed rather than the speed of light.
But we hope that the introduced ideas still might be of interest.The moving electron itself generates an magnetic field, which in turn affects the motion. The generated
field is random, with the probability resulting from summation over all possible intermediate fields.Fix π β₯ 0 called the interaction constant19 (it is related to the electron charge). Fix a black square
(π₯, π¦) called the final position. Fix the rectangle π formed by all the squares (π₯β², π¦β²) such that 1βπ¦ < π₯β² < π¦
17One should think of the value π¦ as fixed, and the quadruples (πΉ, πΉ β², πΈ,πΈβ²) as the possible outcomes of an experiment.18This definition makes sense only for π₯0 β₯ 2π¦; otherwise annihilation of particles cannot be ignored, and the model
becomes inappropruate.19If the lattice step π is not fixed, then π might depend on it. This is called renormalization; we do not discuss it.
7
and 0 < π¦β² < π¦. This waveproof box π encircles any possible checker path π from (0, 0) to (π₯, π¦) excludingthe first and the last move; it is required to make the sum over all intermediate fields below finite.
Take any such path π and any assignment π’ of the numbers Β±1 to all the vertices in the rectangle π .Let π be the number of negative white squares in π for π’ (they play the role of turnings of the checker).Denote by
οΏ½οΏ½(π , π’, π) =ππ
2(π¦β2)2(1 + π2)(π¦β1)2/2π βππ/2οΏ½οΏ½(π , π’)
the vector20 οΏ½οΏ½(π , π’), rotated clockwise through πΒ·90β, multiplied by ππ and divided by 2(π¦β2)2(1+π2)(π¦β1)2/2.Let π’fin be any assignment of the numbers Β±1 to all the vertices on the top side of π such that π’fin
equals +1 at the endpoints of the side. It is called the final magnetic field. Denote by
οΏ½οΏ½(π₯, π¦, π’fin, π,+) :=βπ ,π’
οΏ½οΏ½(π , π’, π)
the sum over all the paths π from (0, 0) to (π₯, π¦) starting and ending with an upwards-right move and overall the assignments π’ of the numbers Β±1 to all the vertices in the rectangle π such that
π’ =
{π’fin on the top side of π ;
+1 on all the other sides of π .
The length square π (π₯, π¦, π’fin, π,+) = |π(π₯, π¦, π’fin, π,+)|2 is the probability that the final magnetic fieldequals π’fin and a right electron is found at (π₯, π¦). Define π (π₯, π¦, π’fin, π,β) analogously.
36. Find the probabilities π (π₯, π¦, π’fin, 1,+) and π (π₯, π¦, π’fin, 1,β) for π=1, π¦=2, 3 and all possible π₯ and π’fin.
The sum21
π (π₯, π¦, π,+) =βπ’fin
π (π₯, π¦, π’fin, π,+)
over all assignments π’fin of the numbers Β±1 to the vertices on the top side of π is the probability to find aright electron at (π₯, π¦). Define π (π₯, π¦, π,β) analogously. For instance, π (π₯, π¦, 0,+) = π (π₯, π¦,+) (why?).
37.* For a path π from (0, 0) to (π₯, π¦), consider the sumβ
π’ οΏ½οΏ½(π , π’, π) over all assignments π’ of Β±1 to allthe vertices in the rectangle π such that π’ = +1 on the boundary of π . Express the sum through οΏ½οΏ½(π )and the number of white squares in each of the two parts, into which π is divided by the path π .
38.* Let π’Β± be equal to Β±1 respectively at the top-right corner of the square (2 β π¦, π¦ β 1), and equal to+1 at all the other vertices on the top side of π . Solve analogues of Problems 4 and 5 for οΏ½οΏ½(π₯, π¦, π’Β±, π,Β±)and π (π₯, π¦, π,+) + π (π₯, π¦, π,β) instead of π1(π₯, π¦), π2(π₯, π¦), and π (π₯, π¦).
9 Creation and annihilation***Question: the same as in the model with identical particles and antiparticles.Assumptions: electron-positron pairs now created and annihilated, no interaction, encircling by reflecting walls.Results: no; this is only an ingredient for more realistic models with interaction.
Finally we have reached an unexplored area: we state an almost 40-years-old open problem.Start with mentioning what is not done in this section:
β we do not give a definition of the new upgrade (it is unknown so far);
β the upgrade (even if defined) would not explain any new experimental results.
But
β we do give a precise statement of the problem: which exactly properties of the upgrade are requested;
20It is well-defined because the path π is contained in π except the first and the last move. We set ππ = 1 for π = π = 0.21Here we sum probabilities rather than vectors. The notation π (π₯, π¦, π,+) should not be confused with π (π₯, π¦,ππ,+).
8
β the upgrade is an important ingredient of further ones fantastically agreeing with experiment.
Informally, our plan is as follows. Checker paths turning downwards or upwards or forming cycles meancreation and annihilation of electron-positron pairs. Even if we start with just one electron, we might endup with many electrons and positrons. To each possible configuration of the resulting particles, we wantto assign a complex number so that its length square is the probability of the configuration in a sense. Thenumber itself is the sum over all possible transitions from the initial configuration to the final one, that is,all possible paths configurations joining them. To make the sum finite, we put reflecting walls around.
Fix a rectangle π formed by all the squares (π₯, π¦) such that π₯min β€ π₯ β€ π₯max and 0 β€ π¦ β€ π¦max (thelines π₯ = π₯min and π₯ = π₯max are called reflecting walls). Fix π, π > 0 called mass and lattice step.
An initial configuration is any assignment22 of one of the signs β+β or βββ to some vertices inside π lying between the lines π¦ = 0 and π¦ = 1. Physically the signs mean the initial positions of positrons andelectrons respectively (and points without a sign are vacant). For our game, this means that the checkerspass the vertices with the βββ sign upwards-left or -right, and the vertices with the β+β sign β downwards-left or -right (and do not pass through vertices without a sign). Analogously, a final configuration is anassignment to vertices between the lines π¦ = π¦max and π¦ = π¦max β 1. An intermediate configuration is anyassignment of one of the signs β+β or βββ to some vertices inside π such that the difference between thenumber of β+β and βββ signs on the 2 top vertices of each black square in the strip 1 β€ π¦ β€ π¦maxβ1 equalsthe difference on the 2 bottom vertices. For our game, this means that the checkers start and finish motionin the lines π¦ = 0 or π¦ = π¦max only. In other words, the signs at the vertices of each black square are inone of the 19 positions23 shown in Figure 6 to the left. These 19 positions are called basic configurations.
Figure 6: Basic configurations
Suppose that one has assigned a complex number (depending on ππ) to each of the 19 basic configu-rations. A βrightβ choice of the numbers is unknown; think of them as fixed parameters of our model.
Then take any intermediate configuration. In each black square inside π not having common pointswith the boundary, write the complex number assigned to the basic configuration in the black square.Assign the product of all the written numbers to the intermediate configuration.
Now to any pair of initial and final configurations, sum the complex numbers assigned to all possibleintermediate configurations between them. Assign the resulting complex number to the pair.
39. (Puzzle) Restrict to configurations without β+β signs. Let π΄,π΄β², π΅,π΅β², πΈ, πΈ β², πΉ, πΉ β² be the black squaresdefined in Β§7. Take any π₯min < βπ¦ and π₯max > π₯0 + π¦. Fix the initial and final configurations with exactlytwo βββ signs, located at the top-right vertices of the squares π΄, π΄β², and πΈ, πΈ β² respectively. Assign complexnumbers to the 6 basic configurations without β+β signs so that the sum of the numbers assigned to allintermediate configurations without β+β signs equals to the vector οΏ½οΏ½(π΄π΅,π΄β²π΅β² β πΈπΉ,πΈ β²πΉ β²) from Β§7.
A case of particular interest is when the initial configuration consists of just one βββ sign at the top-right vertex of the square (0, 0) and the final configuration consists of just one βββ sign at the bottom-right
22The assignment has nothing to do with the magnetic field from Β§6.23We consider positions of signs, but not possible checker paths in a particular black square like in Figure 6 to the right.
9
or bottom-left vertex of a black square (π₯, π¦max). The complex numbers assigned to the pairs in questionare denoted by οΏ½οΏ½(π₯, π¦max,ππ, π₯min, π₯max,β) and οΏ½οΏ½(π₯, π¦max,ππ, π₯min, π₯max,+) respectively.
The desired continuum limit of these complex numbers involves the following modified Bessel functionsand Hankel functions :
πΎ0(π₯) =
β« β
1
πβπ₯π‘
βπ‘2 β 1
ππ‘ π»(1)0 (π₯) =
2
ππ
β« β
1
πππ₯π‘βπ‘2 β 1
ππ‘
πΎ1(π₯) = π₯
β« β
1
πβπ₯π‘βπ‘2 β 1 ππ‘ π»
(1)1 (π₯) = β2π₯
ππ
β« β
1
πππ₯π‘βπ‘2 β 1 ππ‘
40.*** Continuum limit. Assign complex numbers to the 19 basic configurations so that for each |π¦| < |π₯|
limπββ
limπ₯maxββ
π οΏ½οΏ½(
2βππ₯
2
β, 2βππ¦
2
β,π
π,βπ₯max, π₯max,β
)= ππΎ0(π
βπ₯2 β π¦2);
limπββ
limπ₯maxββ
π οΏ½οΏ½(
2βππ₯
2
β, 2βππ¦
2
β,π
π,βπ₯max, π₯max,+
)= βππ
π₯ + π¦βπ₯2 β π¦2
πΎ1(πβπ₯2 β π¦2);
and for each |π¦| > |π₯| we have
limπββ
limπ₯maxββ
π οΏ½οΏ½(
2βππ₯
2
β, 2βππ¦
2
β,π
π,βπ₯max, π₯max,β
)= ππ
π
2π»
(1)0 (π
βπ¦2 β π₯2);
limπββ
limπ₯maxββ
π οΏ½οΏ½(
2βππ₯
2
β, 2βππ¦
2
β,π
π,βπ₯max, π₯max,+
)= π
π
2Β· π₯ + π¦β
π¦2 β π₯2π»
(1)1 (π
βπ¦2 β π₯2).
Let us discuss the physical meaning of the upgrade. For π¦ β« |π₯| the right-hand sides of the lattertwo equations are very close to the right-hand sides of the answer to Problem 10. Thus one may wish tointerpret the upgrade as a more accurate approximation for the probability to find the electron in a square(π₯, π¦). But this faces serious objections.
First, it is in principle impossible to measure the coordinates of an electron exactly 24. Such a prioriuncertainty has the same order of magnitude as the correction introduced by the upgrade. Thus theupgrade does not actually add anything to description of the electron motion.
Second, for fixed initial configuration, the squares of the absolute values of the numbers assigned toall the possible final configurations do not sum up to 1 (even in the continuum limit). The reason is thatdistinct configurations are not mutually exclusive: for a field in a given configuration, there is a positiveprobability to find it in a different configuration. We do not know any clear explanation of that.
To summarize, the upgrade lacks a direct physical interpretation, and should be considered as aningredient for further upgrades.
10 (1 + 1)-dimensional quantum electrodynamics***
Question: what is the probability to find electrons (or electron+positron) with momenta π and πβ² in the far future,if they were emitted with momenta π and πβ² in the far past?Assumptions: interaction now switched on; all simplifying assumptions removed except the default ones:no nuclear forces, no gravitation, electron moves only along the π₯-axis (and π¦-coordinate is interpreted as time).Results: quantum corrections.
Unifying the (so far unknown) upgrades discussed in the previous two sections would give an elementarydefinition of (1 + 1)-dimensional QED.
Future research
An algorithmic quantum field theory is a one which for each experimentally observable quantity and apositive number π provides a precise statement of an algorithm giving the predicted value of the quantitywithin accuracy π. (Surely, the predicted value does not have to agree with the experiment for π less thanaccuracy of theory itself.) This is an extension of constructive quantum field theory, the latter currentlybeing far away from algorithmic one.
24This should not be confused with uncertainty principle, which does not allow simultaneous measurement of the coordi-nates and momentum.
10
Epilogue (underwater rocks)
We hope that at least some of our readers have become interested in elementary particles and want tolearn more about them. As an epilogue, let us give a few warnings to such readers.
In popular science, theory of elementary particles is usually oversimplified. This sequence of problemsis not an exception. The toy models introduced here are very rough and should be considered with agrain of salt. Simplicity is their only advantage; if taken too seriously, the models could even give a wrongphysical intuition. Real understanding of particles theory requires excellent knowledge of both physics andmathematics.
We should also remark that nowadays there are almost no mathematical results in lattice quantumfield theory; what we have is usually just a numeric simulation. Finally, there are βtheories of NewPhysicsβ which are developed without any objective truth criterion: such theories are supported by neitherexperimental nor mathematical proofs (and some of them have experimental disproofs).
11
Hints, solutions, answers
For any vector οΏ½οΏ½ β R2 denote by π1, π2 the coordinates of οΏ½οΏ½. E.g., οΏ½οΏ½(π₯, π¦,ππ,β) = (π1(π₯, π¦,ππ,β), π2(π₯, π¦,ππ,β)).Sometimes the vector οΏ½οΏ½ is considered as a complex number π1 + ππ2 (although complex numbers are not requiredfor the solution of most problems). In what follows assume that π₯ and π¦ have the same parity unless the oppositeis indicated.
1. Answer. In the table, the vector in the cell (π₯, π¦) is οΏ½οΏ½(π₯, π¦), and an empty cell (π₯, π¦) means that οΏ½οΏ½(π₯, π¦) = (0, 0).
4(
12β2, 0) (
0,β 12β2
) (1
2β2,β 1β
2
) (0, 1
2β2
)3
(12 , 0
) (12 ,β
12
) (0, 12
)2
(1β2, 0) (
0, 1β2
)1 (0, 1)
y x β2 β1 0 1 2 3 4
In the following table, the number in the cell (π₯, π¦) is π (π₯, π¦), and an empty cell (π₯, π¦) means that π (π₯, π¦) = 0.
4 1/8 1/8 5/8 1/8
3 1/4 1/2 1/4
2 1/2 1/2
1 1
y x β2 β1 0 1 2 3 4
(The vectors οΏ½οΏ½(π₯, π¦) can be easily computed consecutively using Problem 4.) See also the following figure.
For any positive π¦ we haveβπ₯βZ
π (π₯, π¦) = 1. (This is Problem 5.)
For β4 β€ βπ¦ < π₯ β€ π¦ β€ 4 the probability π (π₯, π¦) equals 0 if and only if one of the numbers π₯, π¦ is odd and theother one is even. For any π₯ > π¦ and π₯ < βπ¦ + 1 we have π (π₯, π¦) = 0. (In general it is not known, if π (π₯, π¦) = 0can vanish for π₯ and π¦ of the same parity satisfying βπ¦ < π₯ β€ π¦.)
For each odd π¦ we have οΏ½οΏ½(0, π¦) = 0. For each nonnegative integer π we have
the vector οΏ½οΏ½(0, 8π+ 2) is directed to the right,
the vector οΏ½οΏ½(0, 8π+ 4) is directed downwards,
the vector οΏ½οΏ½(0, 8π+ 6) is directed to the left,
the vector οΏ½οΏ½(0, 8π+ 8) is directed upwards.
Analogously, for each even π¦ we have οΏ½οΏ½(1, π¦) = 0. For each nonnegative integer π we have
the vector οΏ½οΏ½(1, 8π+ 3) is directed downwards-right,
the vector οΏ½οΏ½(1, 8π+ 5) is directed downwards-left,
the vector οΏ½οΏ½(1, 8π+ 7) is directed upwards-left,
the vector οΏ½οΏ½(1, 8π+ 9) is directed upwards-right.
(See the remark after the answer to Problem 9.)
12
Remark. In the table, 2(π+πβ2)/2οΏ½οΏ½(πβπ+1, π+ πβ 1) stands at the intersection of π-th column and π-th row:
(0, 1) (0, 1) (0, 1) (0, 1) (0, 1) (0, 1) (0, 1) (0, 1)(1, 0) (1,β1) (1,β2) (1,β3) (1,β4) (1,β5) (1,β6) (1,β7)(1, 0) (0,β1) (β1,β1) (β2, 0) (β3, 2) (β4, 5) (β5, 9) (β6, 14)(1, 0) (β1,β1) (β2, 0) (β2, 2) (β1, 4) (1, 5) (4, 4) (8, 0)(1, 0) (β2,β1) (β2, 1) (0, 3) (3, 3) (6, 0) (8,β6) (8,β14)(1, 0) (β3,β1) (β1, 2) (3, 3) (6, 0) (6,β6) (2,β12) (β6,β14)(1, 0) (β4,β1) (1, 3) (6, 2) (6,β4) (0,β10) (β10,β10) (β20, 0)(1, 0) (β5,β1) (4, 4) (8, 0) (2,β8) (β10,β10) (β20, 0) (β20, 20)
2. Answer: no for both questions. For example, π (2, 4) = 5/8 = 1/8+1/4 = π (2, 4 bypass 2, 2)+π (2, 4 bypass 0, 2)and π (0, 4) = 1/8 < 1/4 = π (0, 4 bypass 2, 2).
3. Answer: π (0, 12) = 25/512. Hint: the answer is obtained immediately by means of Remark after the answerto Problem 9 or it can be quickly computed recursively by means of the answer to Problem 4.
4. Answer:
{π1(π₯, π¦) = 1β
2π2(π₯+ 1, π¦ β 1) + 1β
2π1(π₯+ 1, π¦ β 1);
π2(π₯, π¦) = 1β2π2(π₯β 1, π¦ β 1)β 1β
2π1(π₯β 1, π¦ β 1).
Solution. Let us derive the formula for π2(π₯, π¦); the derivation for π1(π₯, π¦) is analogous. Consider anypath π from (0, 0) to (π₯, π¦). Denote by π‘(π ) the number of turnings in π . Denote οΏ½οΏ½(π ) = (π1(π ), π2(π )). Noticethat the component π2(π ) = 0 if and only if π‘(π ) is even and π1(π ) = 0 if and only if π‘(π ) is odd. Therefore,π2(π₯, π¦) =
βπ :π‘(π ) even
π2(π ) and π1(π₯, π¦) =β
π :π‘(π ) odd
π1(π ).
The last move in the path π is made either from (π₯β 1, π¦) or from (π₯+ 1, π¦). It is obvious that if π‘(π ) is even,then the last move is directed upwards-right, else it is directed upwards-left. Since we are interested in π2(π₯, π¦),assume that last move is directed upwards-right.
Denote by π β² the path π without the last move. If the directions of the last moves in π and π β² coincide, thenοΏ½οΏ½(π ) = 1β
2οΏ½οΏ½(π β²), otherwise οΏ½οΏ½(π ) = 1β
2(π2(π
β²),βπ1(π β²)). Therefore,
π2(π₯, π¦) =β
π :π‘(π ) even
π2(π ) =1β2
ββ βπ β²:π‘(π β²) even
π2(π β²)β
βπ β²:π‘(π β²) odd
π1(π β²)
ββ =1β2(π2(π₯β 1, π¦ β 1)β π1(π₯β 1, π¦ β 1)).
5. Let us prove by induction over π¦ thatβπ₯βZ
π (π₯, π¦) = 1 for all π¦ β₯ 1: Obviously,βπ₯βZ
π (π₯, 1) = 1. The step of
induction follows immediately from the following computation:βπ₯βZ
π (π₯, π¦ + 1) =βπ₯βZ
[π1(π₯, π¦ + 1)2 + π2(π₯, π¦ + 1)2
]=
βπ₯βZ
π1(π₯, π¦ + 1)2 +βπ₯βZ
π2(π₯, π¦ + 1)2 =
=1
2
βπ₯βZ
(π1(π₯+ 1, π¦) + π2(π₯+ 1, π¦))2 +1
2
βπ₯βZ
(π2(π₯β 1, π¦)β π1(π₯β 1, π¦))2 =
=1
2
βπ₯βZ
(π1(π₯, π¦) + π2(π₯, π¦))2 +
1
2
βπ₯βZ
(π2(π₯, π¦)β π1(π₯, π¦))2 =
βπ₯βZ
[π1(π₯, π¦)
2 + π2(π₯, π¦)2]=
βπ₯βZ
π (π₯, π¦).
A generalization of conservation law by Gleb Minaev and Ivan Russkikh, participants of Summer conference of
Tournament of towns. Perform the change coordinates (π₯, π¦) β¦β (π‘, π’) = (π₯+π¦2 β 1, π¦βπ₯
2 ), i.e., rotate the coordinatesystem through 45 degrees clockwise and shift it by the vector (β1, 0).
Denote οΏ½οΏ½(π‘, π’) := οΏ½οΏ½(π‘β π’+ 1, π‘+ π’+ 1), π(π‘, π’) := π (π‘β π’+ 1, π‘+ π’+ 1), οΏ½οΏ½(π‘, π’) := (1 +π2π2)(π‘+π’)/2 οΏ½οΏ½(π‘, π’).The coordinates of the vectors οΏ½οΏ½(π‘, π’) and οΏ½οΏ½(π‘, π’) denote by π1(π‘, π’), π2(π‘, π’), π΅1(π‘, π’), π΅2(π‘, π’).
Remark. In the new coordinate system, a checker moves between neighbouring points of the integer latticerather than along the diagonals between black squares. Also we suppose that we start at (0, 0) and move to anypoint of (Nβͺ{0})2, and the additional βpre-moveβ from (β1, 0) to (0, 0) is taken into account only to compute thenumber of turns.
For a subset π β (N βͺ {0})2 denote οΏ½οΏ½(π‘, π’ bypass π) :=β
π οΏ½οΏ½(π ), where we sum is over all paths π from
(0, 0) to (π‘, π’), which bypass the points of the set π . Analogously, define π(π‘, π’ bypass π), οΏ½οΏ½(π‘, π’ bypass π,Β±),π(π‘, π’ bypass π,Β±). Also for a set π denote
π(π) :=βπβπ
π(π bypass π β {π}).
13
Remark. Note that for an infinite set π the sum becomes infinite as well. The order of a summation isirrelevant because all the summands are positive.
Theorem 1. For each finite set π β (N βͺ {0})2 such that there are no infinite paths from (0; 0) bypassing the
points of π we have π(π) = 1.
Proof. Prove the theorem by induction overπ := max(π‘,π’)βπ (π‘+π’), i.e., the maximal number of a downwards-rightdiagonal containing at least one point of the set π . The diagonal is called maximal.
Base: π = 0. In this case π = (0, 0) and π(π) = π(0, 0) = 1.Step. Let us prove a lemma.
Lemma 1. If a set π΄ β (Nβ©{0})2, which does not contain the points (π‘, π’), (π‘, π’+1), (π‘+1, π’) β (Nβ©{0})2 , then
π(π‘, π’ bypass π΄) = π(π‘, π’+ 1 bypass π΄,β) +π(π‘+ 1, π’ bypass π΄,+).
Proof. This is straightforward:
π(π‘, π’+ 1 bypass π΄,β) +π(π‘+ 1, π’ bypass π΄,+) = π1(π‘, π’+ 1 bypass π΄)2 + π2(π‘+ 1, π’ bypass π΄)2 =
=(π1(π‘, π’ bypass π΄) + π2(π‘, π’ bypass π΄))2 + (π2(π‘, π’ bypass π΄)β π1(π‘, π’ bypass π΄))2
2=
= π1(π‘, π’ bypass π΄)2 + π2(π‘, π’ bypass π΄)2 = π(π‘, π’ bypass π΄).
Suppose that the point (π‘, π’) β π is such that π‘+ π’ is maximal. Suppose that there is a checker path startingat the point (0, 0) and ending at (π‘, π’) with the last move, say, in the upwards direction bypassing all the pointsof the set π β {(π‘, π’)}. Then there exists a path to the point (π‘, π’ β 1), bypassing all the points of the setπ β {(π‘, π’ β 1)}. Notice that in this case (π‘, π’ β 1) /β π , because the move to (π‘;π’) is in the upwards direction.Hence there exists a path to the point (π‘ + 1, π’ β 1), bypassing all the points of the set π β {(π‘ + 1, π’ β 1)}. If(π‘ + 1, π’ β 1) /β π then there exists a path going to infinity passing through (π‘ + 1, π’ β 1) and bypassing all thepoints of the set π . For example, the path turning right at the point (π‘, π’) and only going right from there passesthrough (π‘, π’) and bypasses all the points in π because the diagonal is maximal. Therefore (π‘ + 1, π’ β 1) β π .Notice that π(π, π bypass π΄) = π(π, π bypass π΄,+) + π(π, π bypass π΄,β) for any set π΄ β (N β© {0})2. DenoteπΎ := π β{(π‘, π’); (π‘+1, π’β1)}. Then we have the following chain of equalities (where we use 1 for the set π΄ = πΎ):
π(π) = π(πΎ) +π(π‘, π’ bypass π β {(π‘, π’)}) +π(π‘+ 1, π’β 1 bypass π β {(π‘+ 1, π’β 1)}) == π(πΎ) +π(π‘, π’ bypass π β {(π‘, π’)},+) +π(π‘, π’ bypass π β {(π‘, π’)},β)+
+π(π‘+ 1, π’β 1 bypass π β {(π‘+ 1, π’β 1)},+) +π(π‘+ 1, π’β 1 bypass π β {(π‘+ 1, π’β 1)},β) =
= π(πΎ) +π(π‘, π’ bypass π β {(π‘, π’)},+)+
+π(π‘, π’β 1 bypass π β {(π‘, π’β 1)}) +π(π‘+ 1, π’β 1 bypass π β {(π‘+ 1, π’β 1)},β) = π(π βͺ {(π‘, π’β 1)}).
Thus if the point (π‘, π’β 1) is added to the set π then the probability π(π) is not changed.This way we put new points onto the diagonal π‘+ π’ = πβ 1, therefore, the maximal diagonal is not changed.
Thus if we add a few points to the set π , then the paths bypassing other points of the set π bypass the points ofπ in the maximal diagonal as well. Therefore, if we remove all the points from the maximal diagonal, then π(π)is not changed and no infinite path bypassing the points of the set π appears. This way we change π keepingπ(π) fixed but decreasing the number of a maximal diagonal. By the inductive hypothesis the new π(π) equals1, hence the old one also equals 1.
6. Answer. For each π₯ we have π1(π₯, π¦) = π1(βπ₯, π¦) and π2(π₯, π¦) + π1(π₯, π¦) = π2(2β π₯, π¦) + π1(2β π₯, π¦).Solution. For each path π denote by π(π ) the reflection of π with respect to the π¦ axis. Then if π is a path to
(π₯, π¦), then π(π ) is a path to (βπ₯, π¦).For each path π denote by π(π ) the path consisting of the same moves as π , but in the opposite order. Notice
that reordering of moves does not affect the endpoint.Now consider a path π to (π₯, π¦) such that π‘(π ) is odd. Then the last move in π is upwards-left. Therefore,
the last move in π(π ) is upwards-right, hence the first move in π(π(π )) is upwards-right. Thus π β π is a bijectionbetween paths to (π₯, π¦) with odd number of tunings beginning with an upwards-right move and paths to (βπ₯, π¦)with odd number of tunings beginning with an upwards-right move.
To prove the second equation use the result of Problem 4:
π1(π₯, π¦) + π2(π₯, π¦) =β2 Β· π1(π₯β 1, π¦ + 1) =
β2 Β· π1(1β π₯, π¦ + 1) = π1(2β π₯, π¦) + π2(2β π₯, π¦).
14
Remark. These identities can also be proved simultaneously by induction over π¦ using Problem 4. AlsoProblem 9 implies another identity (π¦ β π₯) π2(π₯, π¦) = (π¦ + π₯β 2) π2(2β π₯, π¦).
7. Answer : for each 0 < π¦β² < π¦ we have
π1(π₯, π¦) =βπ₯β²
[π2(π₯
β², π¦β²)π1(π₯β π₯β² + 1, π¦ β π¦β² + 1) + π1(π₯β², π¦β²)π2(π₯
β² β π₯+ 1, π¦ β π¦β² + 1)],
π2(π₯, π¦) =βπ₯β²
[π2(π₯
β², π¦β²)π2(π₯β π₯β² + 1, π¦ β π¦β² + 1)β π1(π₯β², π¦β²)π1(π₯
β² β π₯+ 1, π¦ β π¦β² + 1)].
The required expression for οΏ½οΏ½(π₯, 199) is obtained by taking π¦ = 199 and π¦β² = 100.Solution. Fix any positive integer π¦β² < π¦. Consider a path π from (0, 0) to (π₯, π¦). Denote by (π₯β², π¦β²) the square
at which π intersects the line π¦ = π¦β². Denote by π 1 the part of π that joins (0, 0) with (π₯β², π¦β²) and by π 2 the partstarting at the intersection square of π with the line π¦ = π¦β² β 1 and ending at (π₯, π¦). Translate the path π 2 so thatit starts at (0, 0).
Denote οΏ½οΏ½(π ) = (π1(π ), π2(π )). Then
π1(π ) =
{π1(π 1)π2(π 2) if the move to (π₯β², π¦β²) is upwards-left,
π2(π 1)π1(π 2) if the move to (π₯β², π¦β²) is upwards-right,
π2(π ) =
{βπ1(π 1)π1(π 2), if the move to (π₯β², π¦β²) is upwards-left,
π2(π 1)π2(π 2), if the move to (π₯β², π¦β²) is upwards-right.
Therefore,
π1(π₯, π¦) =βπ
π1(π ) =βπ₯β²
βπ through (π₯β²,π¦β²)
π1(π ) =βπ₯β²
β‘β’β’β’β’β’β£β
π through (π₯β²,π¦β²),move to (π₯β²,π¦β²)is upwards-right
π2(π 1)π1(π 2) +β
π through(π₯β²,π¦β²),move to (π₯β²,π¦β²)is upwards-left
π1(π 1)π2(π 2)
β€β₯β₯β₯β₯β₯β¦ =
=βπ₯β²
[π2(π₯
β², π¦β²)π1(π₯β π₯β² + 1, π¦ β π¦β² + 1) + π1(π₯β², π¦β²)π2(π₯
β² β π₯+ 1, π¦ β π¦β² + 1)].
The formula for π2(π₯, π¦) is proven analogously.Remark. Formulae illustrating the Huygens principle have the form of a convolution. In combinatorics the
convolution of two given sequences {ππ} and {ππ} (π, π β₯ 0) is the sequence {ππ} (π β₯ 0), defined by the equalityππ =
βπ+π=π ππππ. A convolution of two sequences corresponds to the product of their generating functions: if
π΄(π‘) =
ββπ=0
πππ‘π, π΅(π‘) =
ββπ=0
πππ‘π, πΆ(π‘) =
ββπ=0
πππ‘π,
then πΆ(π‘) = π΄(π‘)π΅(π‘). Therefore, choosing the generating functions in a right way, we can write down the Huygensprinciple in a more compact form.
Define two series of Laurent polynomials (i.e., polynomials in variables π‘ and π‘β1):
ππ = ππ(π‘) =
πβπ=βπ+2
π1(π,π)π‘π, ππ = ππ(π‘) =
πβπ=βπ+2
π2(π,π)π‘π.
15
These polynomials can be considered as generating fuctions of the sequences π1,2(π,π) with fixed π. From theanswer to the Problems 6 and 7 we have
π2(π₯, π¦) =βπ₯β²
[π2(π₯
β², π¦β²)π2(π₯β π₯β² + 1, π¦ β π¦β² + 1)β π1(π₯β², π¦β²)π1(π₯
β² β π₯+ 1, π¦ β π¦β² + 1)]=
=βπ₯β²
[π2(π₯
β², π¦β²)π2(π₯β π₯β² + 1, π¦ β π¦β² + 1)β π1(π₯β², π¦β²)π1(π₯β π₯β² β 1, π¦ β π¦β² + 1)
].
This formula is equivalent to the following addition theorem for πβpolynomials:
ππ+π =1
π‘ππππ+1 β π‘ππππ+1.
Also from the Problems 6 and 7 we have
π1(π₯, π¦) =βπ₯β²
[π2(π₯
β², π¦β²)π1(π₯β π₯β² + 1, π¦ β π¦β² + 1) + π1(π₯β², π¦β²)π2(π₯
β² β π₯+ 1, π¦ β π¦β² + 1)]=
=βπ₯β²
[π2(π₯β², π¦β²)π1(π₯β π₯β² + 1, π¦ β π¦β² + 1)+
+ π1(π₯β², π¦β²)(π2(π₯β π₯β² + 1, π¦ β π¦β² + 1) + π1(π₯β π₯β² + 1, π¦ β π¦β² + 1)β π1(π₯β π₯β² β 1, π¦ β π¦β² + 1))].
This formula respectively is equivalent to the following addition theorem for πβpolynomials:
ππ+π =1
π‘(ππππ+1 + ππππ+1) +
(1
π‘β π‘
)ππππ+1.
In particular substitution of π = 1 gives Diracβs equation written in terms of πβ and πβpolynomials:{ππ+1 = 1
π‘πππ2 + ππ Β· (1π‘π2 +1π‘π2 β π‘π2) =
1π‘β2(ππ + ππ),
ππ+1 = 1π‘πππ2 β π‘πππ2 =
π‘β2(ππ β ππ).
It follows immediately that{π‘β2ππ+1 βππ = ππ = ππ β
β2π‘ ππ+1
π‘β2ππ+1 β ππ = ππ = ππ +
β2π‘ ππ+1
β
{ππ+1 =
β2π‘ ππ β ππ+1
π‘2
ππ+1 = π‘2ππ+1 β π‘β2ππ
β
{ππ+1 = 1β
2(π‘+ 1
π‘ )ππ β ππβ1
ππ+1 = 1β2(π‘+ 1
π‘ )ππ βππβ1.
The latter formulae give us yet another recursive definitions for ππ and ππ.
8. Answer : the graphs of π (π₯, 1000), π1(π₯, 1000), and π2(π₯, 1000) respectively as functions in π₯:
-1000 -500 500 1000
0.0005
0.0010
0.0015
0.0020
0.0025
0.0030
-1000 -500 500 1000
-0.06
-0.04
-0.02
0.02
0.04
0.06
-1000 -500 500 1000
-0.05
0.05
In the above graph of π (π₯, 1000), large values are cut out, to make the oscillations more visible. Compare withthe following graphs of π (π₯, 1000) and π (π₯, π¦) without cutting (kindly provided by A. Daniyarkhodzhaev andF. Kuyanov, participants of the Summer School in Contemporary Mathematics in Dubna, 2019):
16
9. Answer 1: for each π¦ > |π₯| we have
π1(π₯, π¦) = 2(1βπ¦)/2β
π€β‘2π¦βπ₯β2
2
(β1)π¦βπ₯β2+2π€Β·sign(π₯)
4
(|π₯|π€
)( π¦β|π₯|β22
π¦βπ₯β2β2π€4
),
π2(π₯, π¦) = 2(1βπ¦)/2β
π€β‘2π¦βπ₯2
(β1)π¦βπ₯+2π€Β·sign(π₯)
4
(|π₯|π€
)( π¦β|π₯|β22
π¦βπ₯β2π€4
);
for each π¦ = π₯ > 0 we have
π1(π¦, π¦) = 0,
π2(π¦, π¦) = 2(1βπ¦)/2;
for each 0 < π¦ < |π₯| or π¦ = βπ₯ > 0 we have
π1(π₯, π¦) = 0,
π2(π₯, π¦) = 0.
Answer 2: for each π¦ > |π₯| we have
π1(π₯, π¦) = 2(1βπ¦)/2
βπ¦/2ββπ=0
(β1)π((π₯+ π¦ β 2)/2
π
)((π¦ β π₯β 2)/2
π
),
π2(π₯, π¦) = 2(1βπ¦)/2
βπ¦/2ββπ=1
(β1)π((π₯+ π¦ β 2)/2
π
)((π¦ β π₯β 2)/2
π β 1
);
For other π¦ the answer is the same as Answer 1.Answer 3. For even π+ π and 2β π β€ π β€ π we have25
π1(π, π) = (β1)π2(πβ3)/2 Β·πβ2β
π=π+|π|β2
2
(β2)βπ Β· π!
(π β 2β π)!(πβπ2 + 1 + π)!(π β π+π
2 + 1)!
π2(π, π) = 2(1βπ)/2 Β·(π β 1πβπ2
)β (β1)π2(πβ3)/2
πβ2βπ=
π+|π|β22
(β2)βπ Β· π!(π + π+ 1)
(π β 1β π)!(πβπ2 + 1 + π)!(π β π+π
2 + 1)!
Remark. Answer 2 shows that π1(βπ¦+2π, π¦) and π2(βπ¦+2π, π¦) are the coefficients before π‘π¦βπβ1 and π‘π¦βπ inthe expansion of 2(1βπ¦)/2(1 + π‘)π¦βπβ1(1β π‘)πβ1 for 1 β€ π β€ π¦ β 1. In particular,
π1(0, 4π+ 2) =(β1)π
2(4π+1)/2
(2π
π
),
π2(0, 4π+ 2) = 0,
and
π1(0, 4π) = 0,
π2(0, 4π) =(β1)π
2(4πβ1)/2
(2πβ 1
π
).
The sums in the expressions for π1(π₯, π¦) and π2(π₯, π¦) for π¦ > |π₯| are particular cases of well-known Gaussianhypergeometric functions (but familiarity with them is not required for what follows):
π1(π₯, π¦) = 2(1βπ¦)/22πΉ1
(1β π₯+ π¦
2, 1 +
π₯β π¦
2, 1;β1
),
π2(π₯, π¦) = 2(1βπ¦)/2
(1β π₯+ π¦
2
)2πΉ1
(2β π₯+ π¦
2, 1 +
π₯β π¦
2, 2;β1
).
Hint. Let us find π1(π₯, π¦) for π¦ > |π₯|. Consider a path from the square (0, 0) to the square (π₯, π¦) with anodd number of turns (the others do not contribute to π1(π₯, π¦)). Denote by π and πΏ the number of upwards-right
25Fractions in those equalities are respectively(π
π β 2β π, πβπ2 + 1 + π, π β π+π
2 + 1
)and
(π
π β 2β π, πβπ2 + 1 + π, π β π+π
2 + 1
)Β· π + π+ 1
π β 1β π.
17
and upwards-left moves respectively. These numbers are uniquely determined by the conditions πΏ + π = π¦ andπ β πΏ = π₯. Denote by 2π + 1 the number of turns in the path. Let π₯1, π₯2, . . . , π₯π+1 be the number of consecutiveupwards-right moves before the first, the third, . . . , the last turn respectively. Let π¦1, π¦2, . . . , π¦π+1 be the numberof consecutive upwards-left moves after the first, the third, . . . , the last turn respectively. Then π₯π, π¦π β₯ 1 and
π = π₯1 + Β· Β· Β·+ π₯π+1;
πΏ = π¦1 + Β· Β· Β·+ π¦π+1.
Conversely, each collection (π₯1, π₯2, . . . , π₯π+1, π¦1, π¦2, . . . , π¦π+1) satisfying the resulting equations determines a pathfrom (0, 0) to (π₯, π¦) with an odd number of turns.
The problem now reduces to a combinatorial one: find the number of positive integer solutions of the resultingequations. For the first equation, this number equals to the number of ways to put π sticks between π coins in arow, that is,
(π β1π
). Thus
π1(π₯, π¦) = 2(1βπ¦)/2
βπ¦/2ββπ=0
(β1)π(π β 1
π
)(πΏβ 1
π
).
Since πΏ+π = π¦ and π β πΏ = π₯, the required formula from Answer 2 follows.The formula for π1(π₯, π¦) from Answer 1 can be derived using the polynomial from the remark after the answer.
Depending on the sign of π₯ we get
2(1βπ¦)/2(1 + π‘)π¦βπβ1(1β π‘)πβ1 =
{2(1βπ¦)/2(1β π‘2)
π¦βπ₯β22 (1β π‘)π₯, for π₯ β₯ 0;
2(1βπ¦)/2(1β π‘2)π¦+π₯β2
2 (1 + π‘)βπ₯, for π₯ < 0.
For each |π₯| < π¦ the numbers π1(π₯, π¦) and π2(π₯, π¦) are equal to the coefficients before π‘π₯βπ¦β2
2 and π‘π₯βπ¦2 in the
expansion of the polynomials. Considering the cases π₯ β₯ 0 and π₯ < 0 separately we get the required formula.The proof of the formulae for π2(π₯, π¦) for π¦ > |π₯| is analogous. The case π¦ β€ |π₯| is obvious.
Solution by Alexander Kudryavtsev (participant of Summer conference of Tournament of Towns). Denoteππ,π = π1(2π β π + 1, π + 1) Β· 2π/2, ππ,π = π2(2π β π + 1, π + 1) Β· 2π/2 (they are well-defined for 0 β€ π β€ π β 1). LetπΆ(π₯, π¦), π·(π₯, π¦) be generating functions of ππ,π and ππ,π respectively. From the answer to Problem 4 we obtain thefollowing equations: β§βͺβ¨βͺβ©
πΆ(π₯, π¦)β πΆ(π₯, 0)
π¦= π·(π₯, π¦) + πΆ(π₯, π¦)
π·(π₯, π¦)βπ·(π₯, 0)
π¦= π₯ Β· (π·(π₯, π¦)β πΆ(π₯, π¦))
Since πΆ(π₯, 0) = 0 and π·(π₯, 0) = 1, we get{πΆ(π₯, π¦) = π¦ Β· (π·(π₯, π¦) + πΆ(π₯, π¦))
π·(π₯, π¦)β 1 = π₯π¦ Β· (π·(π₯, π¦)β πΆ(π₯, π¦))
The solution of this system is πΆ(π₯, π¦) =π¦
1β π¦ β π₯π¦ + 2π₯π¦2and π·(π₯, π¦) =
1β π¦
1β π¦ β π₯π¦ + 2π₯π¦2. Denote
πΈ(π₯, π¦) :=πΆ(π₯, π¦)
π¦=
π·(π₯, π¦)
1β π¦= 1 + π¦(1 + π₯β 2π₯π¦) + π¦2(1 + π₯β 2π₯π¦)2 + . . .
The coefficient of π₯π Β· π¦π in the πΈ(π₯, π¦) is equal to
πβπ=max(π,πβπ)
(β2)πβπ Β· π!
(πβ π)!(πβπ+ π)!(π β π)!,
since from every summand of the form π¦π(1 + π₯β 2π₯π¦)π we must take exactly one combination of factors:
β for the power of π¦ to be equal to π, the number of factors β2π₯π¦ must be exactly πβ π;
β for the power of π₯ to be equal to π, the number of factors π₯ must be exactly πβ (πβ π);
β for the total number of factors to be π, the number of factors 1 must be πβ (πβ (πβπ))β (πβπ) = πβπ.
18
And the number of ways to choose π β π, π β π and π β (π β π) factors (1 + π₯ β 2π₯π¦) equals to the fraction inthe formula for πΈ(π₯, π¦).
Now Answer 3 can be easily obtained from the equations πΆ(π₯, π¦) = π¦ Β·πΈ(π₯, π¦) and π·(π₯, π¦) = πΈ(π₯, π¦)βπΆ(π₯, π¦).
10. Answer:
οΏ½οΏ½(π₯, π¦) =
β2(sin
(ππ¦4 β π₯2
2π¦
), ππ₯/π¦ cos
(ππ¦4 + π₯(2βπ₯)
2π¦
))βππ¦
+O
(log2 π¦
π¦
)for |π₯| = O(
βπ¦) .
Remark. Compare the graphs of π (π₯, 400), π1(π₯, 400), and π2(π₯, 400) (blue) with the graphs of their approxi-mations in the right-hand side for π¦ = 400 (orange):
-400 -200 200 400
0.001
0.002
0.003
0.004
0.005
0.006
0.007
-400 -200 200 400
-0.05
0.05
0.10
-400 -200 200 400
-0.10
-0.05
0.05
0.10
Hint. Note that by Problem 9
π2(βπ+ 2π + 1, π+ 1) = 2βπ/2[π₯πβ1](β1)π(1 + π₯)πβπβ1(1β π₯)π;
π1(βπ+ 2π + 1, π+ 1) = 2βπ/2[π₯π](β1)π(1 + π₯)πβπβ1(1β π₯)π.
Thus
π2(βπ+ 2π + 1, π+ 1) = 2βπ/2(β1)πβ« 1
0(1 + π2πππ‘)πβπβ1(1β π2πππ‘)ππβ2ππ(πβ1)π‘ππ‘.
For an odd π denote π = π β πβ12 . We have
π2(βπ+ 2π + 1, π+ 1) =ππβ2
β« 1
0(sin(2ππ‘))
πβ12 (tan(ππ‘))ππβ2ππ(πβ1)π‘ππ‘.
Since π << π, we can calculate the integral in the areas (1/4β πΏ, 1/4 + πΏ), (3/4β πΏ, 3/4 + πΏ) with good accuracy.Denote π βͺ
βπ log π, |π| β€ π.
In (1/4β πΏ, 1/4 + πΏ) we have π‘ = 1/4 + π, |π | β€ πΏ,
sin(2ππ‘) = cos(2ππ) = 1β 2π2π2 +π(π4) = πβ2π2π2 +π(π4);
tan(ππ‘) = 1 + 2ππ +π(π2) = π2ππ +π(π2);
πβ2ππ(πβ1)π‘ = π1βππβ2ππ(πβ1)π
and we getππ+1βπ
β2
β« πΏ
βπΏ(πβ2π2π2 +π(π4))
πβ12 (π2ππ +π(π2))ππβ2ππ(πβ1)πππ. (*)
As ππ4 βͺ 1 and ππ2 βͺ 1, i.e. ππΏ4 βͺ 1 and ππΏ2 βͺ 1,
(*) = ππ+12
β2
β« πΏ
βπΏ(πβ(πβ1)π2π2 +π(π4))(π2πππ +π(π2))πβ2ππ(πβ1)πππ.
If we open the brackets we get the summand π(ππ4π2πππ ) = π(ππΏ4π2πππΏ) = π(πβΞ), since we ask ππΏ βͺ log π.Then
(*) = ππ+12
β2
β« πΏ
βπΏ(πβ(πβ1)π2π2+2πππβ2ππ(πβ1)π +π(πβΞ) +π(ππΏ2))ππ.
Now estimate the errorβ« β
πΏπβ(πβ1)π2π2+2πππππ =
β« β
πΏπβ(πβ1)π2(πβ π
π(πβ1))2+ π2
πβ1ππ βͺβ« β
πΏπβ(πβ1)π2 π2
4+ π2
πβ1ππ β€ 1βπβ 1
πβ(πβ1)πΏ2 ,
since π2 β₯ π
π(πβ1) , i.e. πΏ β₯ π/(πβ 1).
19
So the necessary condition is ππΏ2 β« log π. Then we can calculate
(*) = ππ+12
β2
β« β
ββπβ(πβ1)π2π2+2πππβ2ππ(πβ1)πππ+π(πβΞ)+π(ππΏ3) =
ππ+12
β2
β« β
ββπβ(πβ1)π2π2+2πππβ2ππ(πβ1)πππ+π(log2 π/π),
since ππΏ3 βͺ log π Β· πΏ2 βͺ log2 π/π.Finally, from the formula
β«βββ πβππ‘2+ππ‘ππ‘ =
βπ/πππ
2/4π we get
(*) = ππ+12β
2π(πβ 1)π
(2πβ1)β2π(πβ1)πβ1 +π(log2 π/π).
In the area of 3/4 the integral approximately equals π3π+12β2
π(2πβ1)+2π(πβ1)
πβ1 and, since 3π+12 β‘ βπ+1
2 mod 4,
π2(βπ+ 2π + 1, π+ 1) = π2πβ1
β2
π(πβ 1)cos
(π(π+ 1)
4β 2π2
πβ 1+β 2π
πβ 1
)+π(log2 π/π) =
= βπ2πβ1
β2
π(πβ 1)sin
(π(πβ 1)
4β 2π2
πβ 1+
2π
πβ 1
)+π(log2 π/π).
Analogously,
π1(βπ+ 2π + 1, π+ 1) =
β2
π(πβ 1)cos
(π(πβ 1)
4β 2π2
πβ 1
)+π(log2 π/π).
For an even π the answer is
π2(βπ+ 2π + 1, π+ 1) = π2π
β2
π(πβ 2)cos
(π(π+ 1)
4β 2(π2 β 1/4)
πβ 2
)+π(log2 π/π);
π1(βπ+ 2π + 1, π+ 1) =
β2
π(πβ 2)sin
(π(π+ 1)
4β 2(π + 1/2)2
πβ 2
)+π(log2 π/π),
where π = π β π/2.
11. Answer : π (π₯, π¦) = π (π₯, π¦,+) + π (π₯, π¦,β); οΏ½οΏ½(π₯, π¦,+) = (0, π2(π₯, π¦)); οΏ½οΏ½(π₯, π¦,β) = (π1(π₯, π¦), 0).Solution. Note that if a path π from (0, 0) to (π₯, π¦) starts and finishes with an upward-left move, then the checker
changes the direction an odd number of times. Therefore, οΏ½οΏ½(π ) = (Β±21βπ¦2 , 0). Analogously, if a path π finishes with
an upward-right move, then οΏ½οΏ½(π ) = (0,Β±21βπ¦2 ). Hence, οΏ½οΏ½(π₯, π¦,+) = (0, π2(π₯, π¦)) and οΏ½οΏ½(π₯, π¦,β) = (π1(π₯, π¦), 0).
For the probability, the following formula comprehensively explains the answer:
π (π₯, π¦) = |π(π₯, π¦,+) + οΏ½οΏ½(π₯, π¦,β)|2 = π21(π₯, π¦) + π22(π₯, π¦) = π (π₯, π¦,+) + π (π₯, π¦,β).
12. Answer: for each integer π¦ β₯ 2 we have π (π¦,β) = 12
ββπ¦/2ββ1π=0
1(β4)π
(2ππ
). We have also maxπ¦ π (π¦,β) =
π (2,β) = π (3,β) = 1/2 and limπ¦ββ π (π¦,β) = 12β2.
Remark. The graph of the sequence kindly provided by Gleb Minaev and Ivan Russkikh, participants ofSummer Conference of Tournament of Towns:
Solution. Denote
π1(π¦) =βπ₯
π21(π₯, π¦); π2(π¦) =βπ₯
π22(π₯, π¦); π12(π¦) =βπ₯
π1(π₯, π¦)π2(π₯, π¦).
By Problem 5 for each π¦ β₯ 1 we haveπ1(π¦) + π2(π¦) = 1.
20
By the answers to Problem 4, Problem 6 and Problem 7 we have
π1(0, 2π¦) =1β2
βπ₯
π1(π₯, π¦)(π2(π₯, π¦)β π1(π₯, π¦)) + π2(π₯, π¦)(π2(π₯, π¦) + π1(π₯, π¦)) =1β2(π2(π¦) + 2π12(π¦)β π1(π¦)).
By definition and the answer to Problem 4 we have
π1(π¦ + 1)β π2(π¦ + 1) = 2π12(π¦).
Hence,π1(π¦ + 1)β π2(π¦ + 1) = π1(π¦)β π2(π¦) + π1(0, 2π¦)
β2.
Since π1(π¦) + π2(π¦) = 1, we have the recurrence relation π1(π¦ + 1) = π1(π¦) +1β2π1(0, 2π¦).
The remark from the solution of Problem 9 implies by induction that
π1(π¦) =1
2
βπ¦/2ββ1βπ=0
1
(β4)π
(2π
π
).
It only remains to recall that π (π¦,β) = π1(π¦) by Problem 11.Since the sequence
(2ππ
)14π
decreases, it follows that π (π¦,β) < π (3,β) for each π¦ > 3, thus maxπ¦ π (π¦,β) =π (2,β) = π (3,β) = 1/2.
By the Newton binomial theorem we getββ
π=0
(2ππ
)π₯π = 1β
1β4π₯for each π₯ β
[β1
4 ,14
). Setting π₯ = β1
4 we
obtain limπ¦ββ π (π¦,β) = limπ¦ββ12
ββπ=0
(2ππ
) (β1
4
)π= 1
2β2.
Remark. Using the Stirling formula (see Β§2) one can estimate the convergence rate:
|π (π,β)β limπ¦ββ
π (π¦,β)| < 1
2 Β· 4βπ/2β
(2βπ/2ββπ/2β
)<
π
2πβ
2βπ/2β<
1
2βπ.
13. Answer :
π (π₯, π¦, 0,+) =
{1, for π₯ = π¦;
0, for π₯ = π¦.
π (π₯, π¦,β,+) =
{1, for π₯ = 1, π¦ β‘ 1 mod 2;
0, otherwise.
Solution. By the definition, if the number of turns in a path π is nonzero then οΏ½οΏ½(π , 0) = 0. Therefore,οΏ½οΏ½(π₯, π¦, 0,+) = 0 if and only if π₯ = π¦. Hence, we get the answer for π (π₯, π¦, 0,+). The case π₯ = π¦ is obvious.
One cannot prove the answer for π (π₯, π¦,β,+) because it is a definition; but let us provide a motivation forthe definition. If ππ is βvery largeβ, then οΏ½οΏ½(π ,ππ) is βsmallβ unless the checker turns at each move; this gives theformula for π (π₯, π¦,β,+).
Remark. π (π₯, π¦,β,+) = limππββ π (π₯, π¦,ππ,+). Indeed,
limππββ
|π(π ,ππ)| = limππββ
(ππ)π‘(π )
(β1 + (ππ)2)π¦β1
= limππββ
1
(β1 + (ππ)2)π¦β1βπ‘(π )
=
{1, for π‘(π ) = π¦ β 1;
0, otherwise;
where π‘(π ) is the number of turns in the path π . To compute π (π₯, π¦,ππ,+), we take only paths with π‘ β‘ 0 mod 2,hence we get the answer.
14. Answer. Recall the notation introduced at the beginning of section βHints. . . β. We have
π1(π₯, π¦,ππ,β) =1β
1 +π2π2(π1(π₯+ 1, π¦ β 1,ππ,β) +πππ2(π₯+ 1, π¦ β 1,ππ,+)),
π2(π₯, π¦,ππ,+) =1β
1 +π2π2(π2(π₯β 1, π¦ β 1,ππ,+)βπππ1(π₯β 1, π¦ β 1,ππ,β)),
π2(π₯, π¦,ππ,β) = π1(π₯, π¦,ππ,+) = 0.
For each π¦ > |π₯| we have
π1(π₯, π¦,ππ) = (1 +π2π2)(1βπ¦)/2
βπ¦/2ββπ=0
(β1)π((π₯+ π¦ β 2)/2
π
)((π¦ β π₯β 2)/2
π
)(ππ)2π+1,
π2(π₯, π¦,ππ) = (1 +π2π2)(1βπ¦)/2
βπ¦/2ββπ=1
(β1)π((π₯+ π¦ β 2)/2
π
)((π¦ β π₯β 2)/2
π β 1
)(ππ)2π;
21
For each π¦ = π₯ > 0 we have
π1(π¦, π¦,ππ) = 0,
π2(π¦, π¦,ππ) = (1 +π2π2)(1βπ¦)/2;
For each 0 < π¦ < |π₯| or π¦ = βπ₯ > 0 we have
π1(π₯, π¦,ππ) = 0,
π2(π₯, π¦,ππ) = 0.
Hint. First let us solve the analogue of Problem 4 for ππ = 1. Let us derive the formula for π2(π₯, π¦,ππ,+).Consider any path π from (0, 0) to (π₯, π¦).
The last move in the path π is made either from (π₯β 1, π¦) or from (π₯+ 1, π¦). Denote by π β² the path π withoutthe last move. If the directions of the last moves in π and π β² coincide, then οΏ½οΏ½(π ,ππ) = 1β
1+π2π2οΏ½οΏ½(π β²,ππ), otherwise
οΏ½οΏ½(π ,ππ) = ππβ1+π2π2
(π2(π β²,ππ),βπ1(π
β²,ππ)).
Therefore, π2(π₯, π¦,ππ,+) = 1β1+π2π2
(π2(π₯β 1, π¦ β 1,ππ,+)βπππ1(π₯β 1, π¦ β 1,ππ,β)). The formula for
π1(π₯, π¦,ππ,β) is proved analogously.Now let us solve the analogue of Problem 5: we prove that
βπ₯βZ
π (π₯, π¦,ππ) = 1 for all π¦ β₯ 1 by induction over
π¦ using these results. Obviously,βπ₯βZ
π (π₯, 1,ππ) = 1. The step of induction follows immediately from the following
computation:βπ₯βZ
π (π₯, π¦+1,ππ) =βπ₯βZ
[π1(π₯, π¦ + 1,ππ,β)2 + π2(π₯, π¦ + 1,ππ,+)2
]=
βπ₯βZ
π1(π₯, π¦+1,ππ,β)2+βπ₯βZ
π2(π₯, π¦+1,ππ,+)2 =
=1
1 +π2π2
βπ₯βZ
(π1(π₯+1, π¦,ππ,β)+πππ2(π₯+1, π¦,ππ,+))2+1
1 +π2π2
βπ₯βZ
(π2(π₯β1, π¦,ππ,+)βπππ1(π₯β1, π¦,ππ,β))2 =
=1
1 +π2π2
βπ₯βZ
(π1(π₯, π¦,ππ,β) +πππ2(π₯, π¦,ππ,+))2 +1
1 +π2π2
βπ₯βZ
(π2(π₯, π¦,ππ,+)βπππ1(π₯, π¦,ππ,+))2 =
=βπ₯βZ
[π1(π₯, π¦,ππ,β)2 + π2(π₯, π¦,ππ,+)2
]=
βπ₯βZ
π (π₯, π¦,ππ).
Let us now solve the analogue of Problem 9 for ππ = 1. Let us find π1(π₯, π¦) for π¦ > |π₯|. We need to considerpaths with an odd number of turns only. Denote by 2π+1 the number of turns in a path. Denote by π and πΏ thenumber of upwards-right and upwards-left moves respectively. Let π₯1, π₯2, . . . , π₯π+1 be the number of upwards-rightmoves before the first, the third, . . . , the last turn respectively. Let π¦1, π¦2, . . . , π¦π+1 be the number of upwards-leftmoves after the first, the third, . . . , the last turn respectively. Then π₯π, π¦π β₯ 1 for 1 β€ π β€ π + 1 and
π = π₯1 + Β· Β· Β·+ π₯π+1;
πΏ = π¦1 + Β· Β· Β·+ π¦π+1.
The problem now reduces to a combinatorial one: find the number of positive integer solutions of the resultingequations. For the first equation, this number equals to the number of ways to put π sticks between π coins in arow, that is,
(π β1π
). Thus
π1(π₯, π¦,ππ) = (1 +π2π2)(1βπ¦)/2
βπ¦/2ββπ=0
(β1)π(π β 1
π
)(πΏβ 1
π
)(ππ)2π+1.
Since πΏ+π = π¦ and π β πΏ = π₯, the required formula follows. The proof of the formula for π2(π₯, π¦) for π¦ > |π₯| isanalogous. The case π¦ β€ |π₯| is obvious.15. Answer: for |π₯| < π¦ we have
limπββ
π οΏ½οΏ½(2βππ₯2
β, 2βππ¦2
β,π
π,β
)= (ππ½0(π
βπ¦2 β π₯2), 0);
limπββ
π οΏ½οΏ½(2βππ₯2
β, 2βππ¦2
β,π
π,+
)= (0,βπ
π₯+ π¦βπ¦2 β π₯2
π½1(πβπ¦2 β π₯2));
for |π₯| β₯ π¦ both limits vanish.
22
Remark. The limit coincides with the retarded fundamental solution of Diracβs equation, describing motion ofan electron in the plane.
Solution by Ivan Gaidai-Turlov, Timofey Kovalev and Alexey Lvov, participants of Summer conference of
Tournament of towns. Denote π₯π := 2βππ₯2 β, π¦π := 2βππ¦2 β, π΄ := π₯π+π¦π2 , π΅ := π¦πβπ₯π
2 .
Lemma 2. limπββ
(1 + (ππ )2)
π¦πβ12 = 1.
Proof. We have 1 < (1 + (ππ )2)
π¦πβ12 < (1 + (ππ )
2)ππ¦, therefore if π β β then (1 + (ππ )2)ππ¦ = π
β(1 + (ππ )
2)π2π¦ βπβππ2π¦ β 1. Hence by the squeeze theorem we have lim
πββ(1 + (ππ )
2)π¦πβ1
2 = 1.
Lemma 3. limπββ
π(β1)ππΆ ππ΄β1πΆ
ππ΅β1(
ππ )
2π+1 = (β1)π Β· (π₯+π¦2
)π( π¦βπ₯2
)πΒ·π2π+1
(π!)2.
Proof. Clearly, π Β· (β1)ππΆ ππ΄β1πΆ
ππ΅β1(
ππ )
2π+1 = (β1)π Β· (π΄β1)(π΄β2)...(π΄βπ)Β·(π΅β1)(π΅β2)...(π΅βπ)(π!)2
Β· π2π+1
π2π .
Thus limπββ
π΄βππ = lim
πββπ΄π = lim
πβββππ₯
2β+βππ¦
2β
π = limπββ
ππ₯2+ππ¦
2+π(π)
π = π₯+π¦2 . Analogously, lim
πββπ΅βππ = π¦βπ₯
2 .
Therefore,
limπββ
π Β· (β1)ππΆ ππ΄β1πΆ
ππ΅β1(
π
π)2π+1 = lim
πββ(β1)π Β· (π΄β1)...(π΄βπ)Β·(π΅β1)...(π΅βπ)
(π!)2Β· π
2π+1
π2π= (β1)π Β·
(π₯+π¦2 )π(π¦βπ₯
2 )π Β·π2π+1
(π!)2.
For each π we have |(β1)ππΆ ππ΄β1πΆ
ππ΅β1(
ππ )
2π+1| < |(β1)π Β· (π₯+π¦2
)π( π¦βπ₯2
)πΒ·π2π+1
(π!)2| because π΄βπ
π < π΄π < π¦+π₯
2 andπ΅βππ < π΅
π < π¦βπ₯2 .
Lemma 4. The series π Β·ββπ=0
(β1)ππΆ ππ΄β1πΆ
ππ΅β1(
ππ )
2π+1 converges absolutely for all π.
Proof. We have
π Β·ββπ=0
|(β1)ππΆ ππ΄β1πΆ
ππ΅β1(
π
π)2π+1| <
ββπ=0
|(β1)π Β·(π₯+π¦
2 )π(π¦βπ₯2 )π Β·π2π+1
(π!)2| =
=
ββπ=0
π Β·(π₯+π¦
2 )π(π¦βπ₯2 )π Β·π2π
(π!)2<
ββπ=0
π Β·(π₯+π¦
2 )π(π¦βπ₯2 )π Β·π2π
(π!)= π Β· π
π¦2βπ₯2
4π2
.
Hence limπββ
ββπ=0
(β1)ππΆ ππ΄β1πΆ
ππ΅β1(
ππ )
2π+1 = ( limπββ
ββπ=0;2|π
πΆ ππ΄β1πΆ
ππ΅β1(
ππ )
2π+1)β ( limπββ
ββπ=0;2-π
πΆ ππ΄β1πΆ
ππ΅β1(
ππ )
2π+1).
Lemma 5. Suppose ({π0(π)}, {π1(π)} . . . ) is a sequence of nonnegative sequences such that limπββ
ππ(π) = ππ and
ππ(π) < ππ for each π, π. Then limπββ
ββπ=0
ππ(π) =ββπ=0
ππ.
Proof. Denote π :=ββπ=0
ππ. Then for each π we haveββπ=0
ππ(π) < π. Take any π > 0. Take such π thatπβπ=0
ππ > πβ π.
For each 0 β€ π β€ π take ππ such that βπ‘ β₯ ππ : ππ(π‘) > ππ β π2π+1 . Then for all π > max(π0,π1, . . . ,ππ ) we
haveββπ=0
ππ(π) > πβ π. Therefore, limπββ
ββπ=0
ππ(π) = π.
From Lemmas 5 and 3 it follows that
limπββ
π Β·ββ
π=0;2|π
πΆ ππ΄β1πΆ
ππ΅β1(
π
π)2π+1 =
ββπ=0;2|π
(π₯+π¦2 )π(π¦βπ₯
2 )π Β·π2π+1
(π!)2.
Therefore, by Problem 14 and Lemma 2 we have
limπββ
ποΏ½οΏ½(π₯π, π¦π,π
π,β) = lim
πββπ Β·
ββπ=0
π(β1)ππΆ ππ΄β1πΆ
ππ΅β1(
π
π)2π+1 =
= (
ββπ=0;2|π
Β·(π₯+π¦
2 )π(π¦βπ₯2 )π Β·π2π+1
(π!)2)β (
ββπ=0;2-π
Β·(π₯+π¦
2 )π(π¦βπ₯2 )π Β·π2π+1
(π!)2) = π Β· π½0(π
βπ¦2 β π₯2).
23
It can be proven analogously that limπββ
ποΏ½οΏ½(π₯π, π¦π,ππ ,+) = βππ π₯+π¦β
π¦2βπ₯2Β· π½1(π
βπ¦2 β π₯2).
Solution. Denote modified Bessel functions
πΌ0(π§) :=ββπ=0
(π§/2)2π
(π!)2and πΌ1(π§) :=
ββπ=0
(π§/2)2π+1
π!(π + 1)!.
The answer follows from the following stronger theorem.
Theorem 2. Let 0 < π0 β€ 1/2, min{π¦ + π₯, π¦ β π₯} β₯ π0, π§ := πβπ¦2 β π₯2, π β₯ π0 = π0(π§, π0) = 4π3π§/π20. Then
π1
(2βππ₯2
β, 2βππ¦2
β,π
π,β
)=
π
π
(π½0(π§) +π
(log2 π
π0ππΌ0(π§)
)),
π2
(2βππ₯2
β, 2βππ¦2
β,π
π,+
)= βπ
π
π¦ + π₯βπ¦2 β π₯2
(π½1(π§) +π
(log2 π
π0ππΌ1(π§)
)).
Proof. Denote π₯π = 2βππ₯2
β, π¦π = 2
βππ¦2
β. By Problem 9 we have
π1
(π₯π, π¦π,
π
π,β
)=
(1 +
π2
π2
) 1βπ¦π2 π
π
[π¦π/2]βπ=0
(β1)π((π¦π + π₯π β 2)/2
π
)((π¦π β π₯π β 2)/2
π
)(ππ
)2π,
π2
(π₯π, π¦π,
π
π,+
)=
(1 +
π2
π2
) 1βπ¦π2 π
π
[π¦π/2]βπ=1
(β1)π((π¦π + π₯π β 2)/2
π
)((π¦π β π₯π β 2)/2
π β 1
)(ππ
)2πβ1.
Since 1 + π2
π2 β€ ππ2/π2
it follows that(1 + π2
π2
) 1βπ¦π2 β₯ π
π2
π2 (1βπ¦π
2 ) because π¦π β₯ 1. Also 1βπ¦π2 β₯ βπ¦π
2 . Hence,
1 β₯(1 +
π2
π2
) 1βπ¦π2
β₯ ππ2
π2 (βπ¦π2 ) β₯ 1β π2π¦
2π.
Thus(1 + π2
π2
) 1βπ¦π2
= 1 +π(π2π¦π ).
To find the asymptotic form for π1 and π2 consider the π -th partial sums with π = βlog πβ+1 summands. Theinequalities π¦ β₯ π0 and π β₯ 4
π20guarantee that the total number of summands βπ¦π2 β β₯ π because ππ¦
2 β₯βπ β₯ log π+1
for π > 0.For π β₯ π the ratios of consecutive summands in the expressions for π1 and π2 equal respectively(π
π
)2 ((π¦π + π₯π β 2)/2β π)((π¦π β π₯π β 2)/2β π)
(π + 1)2<
(ππ
)2Β· π(π¦ + π₯)
2πΒ· π(π¦ β π₯)
2π=
π§2
4π 2,(π
π
)2 ((π¦π + π₯π β 2)/2β π)((π¦π β π₯π β 2)/2β π + 1)
(π β 1)π<
(ππ
)2Β· π(π¦ + π₯)
2πΒ· π(π¦ β π₯)
2π β 2=
π§2
4π (π β 1).
From the condition π > π3π§+1 it follows that π = βlog πβ + 1 β₯ 3π§ + 1 and π§2
4π 2 < π§2
4π (πβ1) < 12 . Therefore, in
both cases the error term (i.e., the sum over π β₯ π ) is less then the sum of geometric series with ratio 12 . Hence,
π1
(ππ₯, ππ¦,
π
π,β
)=
π
π
(1 +π
(π2π¦
π
))Β·
Β·
[πβ1βπ=0
(β1)π((π¦π + π₯π β 2)/2
π
)((π¦π β π₯π β 2)/2
π
)(ππ
)2π+π
((π¦2 β π₯2)π
(π !)2Β·(π2
)2π)]
,
π2
(ππ₯, ππ¦,
π
π,+
)=
π
π
(1 +π
(π2π¦
π
))Β·
Β·
[πβ1βπ=1
(β1)π((π¦π + π₯π β 2)/2
π
)((π¦π β π₯π β 2)/2
π β 1
)(ππ
)2πβ1+π
((π¦ + π₯)π (π¦ β π₯)πβ1
(π β 1)!π !Β·(π2
)2πβ1)]
.
The error term in the latter formulae are estimated as follows. Since π ! β₯ (π/3)π and π β₯β
π¦2 β π₯2 3πβπ
2 =3βπ
2 π§, it follows that
(π¦2 β π₯2)π
(π !)2Β·(π2
)2πβ€ (π¦2 β π₯2)π
(π )2πΒ·(3π
2
)2π
β€ πβπ β€ 1
π.
24
Now transform binomial coefficients in the following way:(πΌπβ 1
π
)=
(πΌπβ 1) Β· Β· Β· (πΌπβ π)
π!=
(πΌπ)π
π!
(1β 1
πΌπ
)Β· Β· Β·
(1β π
πΌπ
).
If πΌ β₯ π0 and π < π then ππΌπ < π
πΌπ <βπ
ππ0= 1β
ππ0β€ π0
2 Β· 1π0
= 12 . Hence from the inequalities πβ2π₯ β€ 1 β π₯ < πβπ₯
for 0 β€ π₯ β€ 1/2 it follows that(1β 1
πΌπ
)Β· Β· Β·
(1β π
πΌπ
)β₯ π
β2πΌπ π
β4πΌπ Β· Β· Β· π
β2ππΌπ = π
βπ(π+1)πΌπ β₯ π
βπ2
πΌπ β₯ 1β π 2
πΌπβ₯ 1β π 2
π0π.
Therefore,(πΌπβ1
π
)= (πΌπ)π
π!
(1 +π
(π 2
π0π
)).
Denote π = π 2
π0ππΌ0(π§). Replacing
((π¦π+π₯πβ2)/2
π
)and
((π¦πβπ₯πβ2)/2
π
)in the previous expression for π1 we get
π1
(π₯π, π¦π,
π
π,β
)=
π
π
(1 +π
(π2π¦
π
))Β·
[πβ1βπ=0
(β1)π(π2
)2π (π¦2 β π₯2)π
(π!)2+π(π )
]=
=π
π
(1 +π
(π2π¦
π
))Β·
[ ββπ=0
(β1)π(π2
)2π (π¦2 β π₯2)π
(π!)2+π(π )
]=
π
π
(1 +π
(π2π¦
π
))Β· (π½0(π§) +π(π )).
Here we can replace the sum with π summands by an infinite sum because the βtailβ of alternating series withdecreasing absolute value of the summands can be estimated by the first summand:
ββ
π=π
(β1)π(π2
)2π (π¦2 β π₯2)π
(π!)2
β€ (π§/2)2π
(π !)2β€ 1
π
(the latter inequality has been proved before).We have π2π¦π0 β€ π2 Β· max{π¦ + π₯, π¦ β π₯} Β· min{π¦ + π₯, π¦ β π₯} = π2(π¦2 β π₯2) = π§2 β€ 9π§2 β€ π 2, thus
π2π¦π π½0(π§) β€ π 2
π0ππΌ0(π§), hence π1
(π₯π, π¦π,
ππ ,β
)= π
π (π½0(π§) +π(π )) as required.Analogously,
π2
(π₯π, π¦π,
π
π,+
)=
π
π
(1 +π
(π2π¦
π
))Β· π¦ + π₯β
π¦2 β π₯2Β·
[πβ1βπ=1
(β1)π(π2
)2πβ1 (π¦2 β π₯2)2πβ1
2
(π β 1)!π!+π(
π 2
π0ππΌ1(π§))
]=
= βπ
πΒ· π¦ + π₯β
π¦2 β π₯2
(π½1(π§) +π
(π 2
π0ππΌ1(π§)
)).
16. Answer : see the following table, where the pair (π, π)+ in the cell (π₯, π¦) means that οΏ½οΏ½(π₯, π¦, 4, 3,+) = 1β3(π, π),
and (π, π)β means that οΏ½οΏ½(π₯, π¦, 4, 3,β) = 1β3(π, π); an empty cell (π₯, π¦) means that οΏ½οΏ½(π₯, π¦, 4, 3,+) = οΏ½οΏ½(π₯, π¦, 4, 3,β) =
(0, 0).
3 (0, 0)+(β1, 0)β
(0, 1)+(0, 0)β
(0,β2)+(0, 0)β
(0, 2)+(β1, 0)β
(0,β1)+(0, 0)β
2 (0, 0)+(β1, 0)β
(0,β1)+(1, 0)β
(0, 1)+(β1, 0)β
(0,β1)+(0, 0)β
1 (0,β1)+(0, 0)β
(0, 1)+(0, 0)β
(0,β1)+(0, 0)β
y x β3 β2 β1 0 1 2 3 4 5
Hint. Note that we are interested only in even π₯0, thus 2ππ₯0π/π = ππ₯02 is a multiple of π. Analogously to
Problem 1, that means that οΏ½οΏ½(π₯, π¦, 4, 3,+) is parallel to ππ¦ and οΏ½οΏ½(π₯, π¦, 4, 3,β) is parallel to ππ₯ (be careful: this isnot true for an arbitrary π/π).
17. Let us give a more general formula; the solution of the problem is the particular case ππ = 1. Denote
πβ(π₯, π¦) := οΏ½οΏ½(π₯, π¦,ππ, π/π,Ξ,β);
π+(π₯, π¦) := οΏ½οΏ½(π₯, π¦,ππ, π/π,Ξ,+),
25
where the vectors in the right-hand side are defined analogously to οΏ½οΏ½(π₯, π¦, π/π,Ξ,Β±) and οΏ½οΏ½(π₯, π¦,ππ,Β±). (Bewarethat πβ(π₯, π¦) and π+(π₯, π¦) are vector -valued functions; unlike π1(π₯, π¦) and π2(π₯, π¦) they are not coordinates of anyvector.) Then we have the Dirac equation (where π := π π/2 is the counterclockwise rotation through 90β){
πβ(π₯, π¦) = 1β1+π2π2
πβ(π₯+ 1, π¦ β 1) + βπππβ1+π2π2
π+(π₯+ 1, π¦ β 1);
π+(π₯, π¦) = βπππβ1+π2π2
πβ(π₯β 1, π¦ β 1) + 1β1+π2π2
π+(π₯β 1, π¦ β 1).
From Problem 14 we know that this equation holds for Ξ = 1. It can be generalised for any greater Ξ becauseone can change the order of rotations π 2ππ₯0π/π and π = π π/2.
Using this equation, analogously to Problem 14 we getβπ₯βZ
π (π₯, π¦, π/π,Ξ) = 1 for π¦ β₯ 1 by induction over π¦.
18. Since Ξπ (π₯, π¦, π/π,Ξ,+) = |βΞοΏ½οΏ½(π₯, π¦, π/π,Ξ,+)|2, it suffices to prove that
βΞοΏ½οΏ½(π₯, π¦, π/π,Ξ,+) does not
depend on Ξ for Ξ > π¦+ |π₯|. By definitionβΞοΏ½οΏ½(π₯, π¦, π/π,Ξ,+) =
π₯0=Ξβ1βπ₯0=1βΞπ₯0 even
βπ π 2ππ₯0π/ποΏ½οΏ½(π ). But there are no paths
π starting at (π₯0, 0) and ending at (π₯, π¦) for |π₯β π₯0| > π¦, thus increasing Ξ over π¦ + |π₯| does not change the sum.
19. Answer: if π₯1 and π₯2 are both even, then
οΏ½οΏ½(π₯1, π¦, π/π,+) = π (π₯1βπ₯2)2ππ/ποΏ½οΏ½(π₯2, π¦, π/π,+),
οΏ½οΏ½(π₯1, π¦, π/π,β) = π (π₯1βπ₯2)2ππ/ποΏ½οΏ½(π₯2, π¦, π/π,β).
Solution. From the solution of Problem 18 it follows that οΏ½οΏ½(π₯, π¦, π/π,+) =β
π₯0 even
βπ π 2ππ₯0π/ποΏ½οΏ½(π ), where the
second sum is over those paths from (π₯0, 0) to (π₯, π¦), which both start and finish with an upwards-right move. Toeach path π 1 from (π₯0, 0) to (π₯1, π¦) assign the unique path π 2 from (π₯0 + (π₯2 β π₯1), 0) to (π₯2, π¦), obtained from π 1by translation by (π₯1 β π₯2, 0) and vice versa. Thus we sum over the same vectors, but for π₯ = π₯1 they are rotatedthrough 2ππ₯0π/π while for π₯ = π₯2 they are rotated through 2π(π₯0 + (π₯2 β π₯1))π/π, and we are done.
20. Let us give a more general formula; the solution of the problem is the particular case ππ = 1.
Answer. Denote πΈ = arccos(cos(2ππ/π)β
1+π2π2
). Then for even π₯+ π¦ we have:
οΏ½οΏ½(π₯, π¦, π/π,β) =(ππ cos(2πππ₯/π) sin((π¦ β 1)πΈ)β
1 +π2π2 sinπΈ,ππ sin(2πππ₯/π) sin((π¦ β 1)πΈ)β
1 +π2π2 sinπΈ
);
οΏ½οΏ½(π₯, π¦, π/π,+) =(cos(2ππ(π₯β 2)/π) sin(π¦πΈ)β cos(2πππ₯/π) sin((π¦ β 2)πΈ)
2 cos(2ππ/π) sinπΈ,
β sin(2ππ(π₯β 2)/π) sin(π¦πΈ) + sin(2πππ₯/π) sin((π¦ β 2)πΈ)
2 cos(2ππ/π) sinπΈ
);
π (π₯, π¦, π/π,β) =π2π2 sin2((π¦ β 1)πΈ)
π2π2 + sin2(2ππ/π);
π (π₯, π¦, π/π,+) =π2π2 cos2((π¦ β 1)πΈ) + sin2(2ππ/π)
π2π2 + sin2(2ππ/π);
π (π₯, π¦, π/π,β) + π (π₯, π¦, π/π,+) =1.
Remark. In complex form,
οΏ½οΏ½(π₯, π¦, π/π,β) = ππ₯ππ sin((π¦ β 1)πΈ)β1 +π2π2 sinπΈ
;
οΏ½οΏ½(π₯, π¦, π/π,+) = ππ₯β1(sin((π¦ β 1)πΈ) sin(2ππ/π)β
1 +π2π2 sinπΈ+ π cos((π¦ β 1)πΈ)
),
where π = cos(2ππ/π) + π sin(2ππ/π). This is the solution of the Dirac equation (see the solution of Problem 17)with the (quasi)periodic initial condition πβ(π₯, 0) = 0 and π+(π₯, 0) = (β sin 2ππ₯π
π , cos 2ππ₯ππ ). The number πΈ has
physical meaning of energy.Hint. Formulas for οΏ½οΏ½(π₯, π¦, π/π,+) and οΏ½οΏ½(π₯, π¦, π/π,β) can be proven by induction over π¦. The base is obvious;
the inductive step follows from Problem 17. Then π (π₯, π¦, π/π,+) and π (π₯, π¦, π/π,β) are computed directly.Path to solution. Let us view vectors as complex numbers. Then π πΌοΏ½οΏ½ = (cosπΌ + π sinπΌ)π. Denote π =
cos(2ππ/π) + π sin(2ππ/π); then πβ1 = cos(2ππ/π) β π sin(2ππ/π). Denote π(π¦) = πβπ¦οΏ½οΏ½(π¦, π¦, π/π,+) and π(π¦) =πβπ¦οΏ½οΏ½(π¦, π¦, π/π,β). Denoteπ = ππβ
1+π2π2and π = 1β
1+π2π2. By Problem 19 for even π₯+π¦ we have οΏ½οΏ½(π₯, π¦, π/π,+) =
ππ₯π(π¦) and οΏ½οΏ½(π₯, π¦, π/π,β) = ππ₯π(π¦). By Problem 17 we have ππ₯π(π¦) = πππ₯β1π(π¦ β 1) β ππππ₯β1π(π¦ β 1) and
26
ππ₯π(π¦) = πππ₯+1π(π¦ β 1) β ππππ₯+1π(π¦ β 1). Since π = 0, we can divide each side by ππ₯ and obtain the followingsystem: β§βͺβͺβͺβ¨βͺβͺβͺβ©
π(π¦) = ππβ1π(π¦ β 1)β πππβ1π(π¦ β 1) (1)
π(π¦) = πππ(π¦ β 1)β ππππ(π¦ β 1) (2)
π(1) = ππβ1 (3)
π(1) = 0 (4)
There are multiple ways of solving this system (we have thought about generating functions and about diagonalisingthe βtransfer matrixβ) but the one which is probably most accessible (to math high-school students in Russia) istransforming this system into a linear recurrence relation of degree 2. From (1) we have π(π¦ β 1) = 1
π πππ(π¦) βππ ππ(π¦ β 1). From this and (2) we have π(π¦) = π
π ππ2π(π¦) β 1π πππ(π¦ β 1). From this and (1) we have π(π¦) =
π(π + πβ1)π(π¦ β 1)β π(π¦ β 2).The solution of this recurrence equation is π(π) = π1π
π+ + π2π
πβ, where
π+ =π(π + πβ1) +
βπ2(π + πβ1)2 β 4
2= π cos(2ππ/π) + π
βπ2 sin2(2ππ/π) +π2
πβ =π(π + πβ1)β
βπ2(π + πβ1)2 β 4
2= π cos(2ππ/π)β π
βπ2 sin2(2ππ/π) +π2
are the roots of the equation π‘2βπ(π+πβ1)π‘+1 = 0, and π1 and π2 are constants that are found using (3) and (4).Since we are dealing with complex numbers, the notation
βπ2(π + πβ1)2 β 4 is ambiguous. But we can choose
any of the two roots;β
π2(π + πβ1)2 β 4 denotes 2πβ
π2 sin2(2ππ/π) +π2. It follows that
π1 = ππβ1 β ππβπ+ β πβ
and π2 = ππβ1 β ππ+πβ β π+
.
Note that πΈ = arccos(π cos(2ππ/π)). We have π+ = cosπΈ + π sinπΈ and πβ = cosπΈ β π sinπΈ. Thus we have
οΏ½οΏ½(π₯, π¦, π/π,+) = ππ₯π(π¦) = πππ₯(πβ1βππβ)ππ¦+β(πβ1βππ+)ππ¦β
π+βπβ= πππ₯β1 sin(π¦πΈ)
sinπΈ β ππππ₯ sin((π¦β1)πΈ)sinπΈ .
Now we compute π (π₯, π¦, π/π,+) = οΏ½οΏ½(π₯, π¦, π/π,+) Β· οΏ½οΏ½(π₯, π¦, π/π,+). It follows from formulas above that π = πβ1,π+ = πβ, πβ = π+, π1 = π πβππ+
π+βπβand π2 = π πβππβ
πββπ+. From this (with some simplification) we finally have
π (π₯, π¦, π/π,+) =(π+βπβ)2βπ2(ππ¦β1
+ βππ¦β1β )2
(π+βπβ)2= 1β π2 sin2((π¦β1)πΈ)
sin2 πΈ.
Analogously we get οΏ½οΏ½(π₯, π¦, π/π,β) = ππ₯π sin((π¦β1)πΈ)sinπΈ and π (π₯, π¦, π/π,β) = π2 sin2((π¦β1)πΈ)
sin2 πΈ.
21. Solution. Let (0, 0), (1, 1), (π₯2, 2), Β· Β· Β· , (π₯π¦, π¦) be the squares passed by the path π . To define the vectorοΏ½οΏ½(π ,π(π₯)π), start with the vector (0, 1). While the checker moves straightly, the vector is not changed, but if thechecker changes the direction after the π-th move, the vector is rotated through 90β clockwise and multiplied byπ(π₯π)π. In addition, at the very end the vector is divided by
βπ¦β1π=1
β1 +π2(π₯π)π2, where π¦ is the total number
of moves. The final position of the vector is what we denote by οΏ½οΏ½(π ,π(π₯)π). The numbers π1(π₯, π¦,π(π₯)π) andπ2(π₯, π¦,π(π₯)π) are defined analogously to π1(π₯, π¦) and π2(π₯, π¦), only οΏ½οΏ½(π ) is replaced by οΏ½οΏ½(π ,π(π₯)π).
The formulas analogous to the ones from Problem 4 are the following:
π1(π₯0, π¦,π(π₯)π) =1β
1 +π2(π₯0 + 1)π2(π1(π₯0 + 1, π¦,π(π₯)π) +π(π₯0 + 1)π Β· π2(π₯0 + 1, π¦,π(π₯)π)),
π2(π₯0, π¦,π(π₯)π) =1β
1 +π2(π₯0 β 1)π2(π2(π₯0 β 1, π¦,π(π₯)π)βπ(π₯0 β 1)π Β· π1(π₯0 β 1, π¦,π(π₯)π)).
22. Answer : For each π₯, π¦ we haveπ (π₯, π¦,π0(π₯), π/π,+) = 1;π (π₯, π¦,π0(π₯), π/π,β) = 0.
π (π₯, π¦,π1(π₯), π/π,+) =
{1
1.04 = 2526 , for 0 < π₯ < π¦;
1, otherwise;
π (π₯, π¦,π1(π₯), π/π,β) =
{0.041.04 = 1
26 , for β π¦ < π₯ < 0;
0, otherwise;
Solution. Analogously to Problem 13 we have π (π₯, π¦,π0(π₯), π/π,+) = 1, π (π₯, π¦,π0(π₯), π/π,β) = 0 for eachπ₯, π¦.
27
The condition οΏ½οΏ½(π ,π1(π₯)) = 0 means that the checker, if at all changed the direction, turned in one ofthe squares with the coordinates (0, 2), (0, 4), Β· Β· Β· , (0, 2 Β·
[π¦2
]). Thus the number of turns in any path π with
οΏ½οΏ½(π ,π1(π₯)) = 0 is not greater than 1.To compute π (π₯, π¦,π1(π₯), π/π,+) we need to consider only the paths ending with an upwards-right move β
i.e. without turns. If the path π does not contain any of the squares with coordinates (0, 1), (0, 2), . . . , (0, π¦ β 1)(amounting to π₯ β€ 0 or π₯ β₯ π¦) then π (π₯, π¦,π1(π₯), π/π,+) = 1. Otherwise, the length of the vector οΏ½οΏ½(π ,π(π₯))equals 1β
1+0.22.
To compute π (π₯, π¦,π1(π₯), π/π,β), consider paths having exactly one turn in the square with zero π₯-coordinate.If there is such a path π (i.e. βπ¦ < π₯ < 0), then the length of the vector οΏ½οΏ½(π ,π(π₯)) equals 0.2β
1+0.22. Otherwise, it
equals zero.
23. Solution. If οΏ½οΏ½(π ,π2(π₯)) = 0 then the checker changed the direction only in the squares with the π₯-coordinateequal to 1 or πΏ. To compute π (0, 2πΏ,π2(π₯), 16,β) we need to consider only the paths with an odd number ofturns. Since for πΏ > 1 we have π¦ = 2πΏ < 3(πΏβ 1) (for πΏ = 1 the function π(πΏ) is undefined), it follows that thenumber of turns must be exactly 1. Therefore, either the path starts in the square (0, 0) and changes the directionin the square (πΏ,πΏ), or the path starts in the square (2 β 2πΏ, 0) and changes the direction in (1, 2πΏ β 1). The
vectors for these two paths are related by the rotation through the angle π β π(πΏβ1)4 and division by 1.04. Since
the vector for the latter path has square 126 and the one for the former one has square 1
26Β·1.042 , by the law of cosineswe get
π (0, 2πΏ,π2(π₯), 16,β) =1
26
(1 +
1
1.042β 2
1.04cos
π(πΏβ 1)
4
)β 0.074
(1β cos
π(πΏβ 1)
4
).
The graph of this function is shown in the following figure (kindly provided by Punnawith Thuwajit, a participantof Summer Conference of Tournament of Towns).
24. Answer :
π (π, πΏ,β) =0.1664
1.1664 + cot2 2π(πΏβ1)π
; maxπΏ
π (π, πΏ,β) =0.1664
1.1664β 0.14;
π (π, πΏ,+) =1
1 + 0.1664 sin2 2π(πΏβ1)π
; π (π, πΏ,β) + π (π, πΏ,+) = 1.
Hint. Denote π = 0.2 and π = 2ππ (πΏβ 1). For π¦ > 2πΏβ 2 we have
π (0, π¦,π2(π₯), π,β) =
=
π 2ππ
π(2βπ¦) πβ
1 +π2
βββ1 +1
1 +π2π2ππ +
π2
(1 +π2)2(π2ππ)2 + Β· Β· Β·+ 1
π2
(π2
1 +π2π2ππ
)[π¦β2
2(πΏβ1)
]2(πΏβ1)
ββ 2
,
Then, since cos 2π = 1β 2 sin2 π, we have
π (π, πΏ,β) =π2
1 +π2Β·
1β 1
1 +π2π2ππ Β·
ββπ=0
(π2
1 +π2π2ππ
)π2
=
=π2
1 +π2Β·
1β 1
1 +π2π2ππ
1
1β π2
1+π2 π2ππ
2
=π2
1 +π2Β· |1β π2ππ|2
|1β π2
1+π2 π2ππ|2=
=π2(1 +π2)(2β 2 cos(2π))
(1 +π2)2 +π4 β 2π2(1 +π2) cos(2π)=
4π2(1 +π2)
(1 + 2π2)2 + cot2 π=
0.1664
1.1664 + cot2 2ππ (πΏβ 1)
.
28
Therefore, maxπΏ π (π, πΏ,β) = 0.16641.1664 β 0.14. Analogously,
π (π, πΏ,+) =
1
1 +π2Β·
ββπ=0
(π2
1 +π2π2ππ
)π2
=
1
1 +π2Β· 1
1β π2
1+π2 π2ππ
2
=
=1
(1 +π2)2 +π4 β 2(1 +π2)π2 cos(2π)=
1
1 + 4π2(1 +π2) sin2 π=
1
1 + 0.1664 sin2 2ππ (πΏβ 1)
.
Note that
1β 1
1 + 0.1664 sin2 2ππ (πΏβ 1)
=0.1664
1sin2 2π
π(πΏβ1)
+ 0.1664=
0.1664
1.1664 + cot2 2ππ (πΏβ 1)
.
Hence, π (π, πΏ,β) + π (π, πΏ,+) = 1.
25. Answer :(π2 β 1)2
(π2 + 1)2 + 4π2 cot2 2ππΏππ
and π =
(0.3328 + 0.864
β1.04
)Β· π
π,
where π =β1 +ππ/π (the refractive index ; notice that the relation between π and π has changed when we
passed to reflection inside the glass). For the latter π the limit equals
0.1664
1.1664 + cot22ππΏ(1.08+0.08
β1.04)
π
.
Remark. The first formula in the answer is well-known and confirmed by experiment. A remarkable detailedpopular science discussion of Problems 23β25 can be found in [Feynman], and a derivation of the equations on aphysical level of rigor β in [Landafshitz, Chapter X, Β§86, Problem 4].
Path to solution. Consider Diracβs equation from the solution of Problem 17. Search for a solution of the form
(π+(π₯, π¦), πβ(π₯, π¦)) = (π+(π₯), πβ(π₯))πβ2ππππ¦/π.
Use Problem 20 to guess πΒ±(π₯) separately for π₯ β€ 0, 0 β€ π₯ β€ πΏ/π, and π₯ β₯ πΏ/π. Assume for simplicity that 1/πis an odd integer. Abbreviate π3(π₯) =: π(π₯).
First consider the half-plane π₯ β€ 0 (to the left from glass). We need to specify what is meant by a solutionof Diracβs equation in the half-plane. Analogously to Problem 17, by Diracβs equation in a black square (π₯, π¦) wemean the system β§β¨β©πβ(π₯β 1, π¦ + 1) = 1β
1+π(π₯)2π2πβ(π₯, π¦) +
βππ(π₯)πβ1+π(π₯)2π2
π+(π₯, π¦);
π+(π₯+ 1, π¦ + 1) = βππ(π₯)πβ1+π(π₯)2π2
πβ(π₯, π¦) +1β
1+π(π₯)2π2π+(π₯, π¦).
A solution of Diracβs equation in the half-plane π₯ β€ 0 is a pair (π+(π₯, π¦), πβ(π₯, π¦)) satisfying the equation in eachblack square in the half-plane. In particular, πβ(π₯, π¦) is defined for π₯ β€ 0, whereas π+(π₯, π¦) is defined for π₯ β€ 1.Search for a solution of the form (π+(π₯), πβ(π₯)) = (π π2πππ₯π/π, π(π) πβ2πππ₯π/π) for some π(π) β C to be determinedlater, where the formulae for π+(π₯) and πβ(π₯) hold for π₯ β€ 1 and π₯ β€ 0 respectively. (Physically, π+(π₯, π¦) is theincident wave, πβ(π₯, π¦) is the reflected wave, and | limπβ0 π(π)|2 is the reflection probability.)
For the half-plane π₯ β₯ πΏ/π take (π+(π₯), πβ(π₯)) = (π‘(π) π2πππ₯π/π, 0) for some π‘(π) β C to be determined later,where the formulae for π+(π₯) and πβ(π₯) hold for π₯ β₯ πΏ/π+ 1 and π₯ β₯ πΏ/π respectively. (Physically, π+(π₯) is thetransmitted wave, and | limπβ0 π‘(π)|2 is the transmission probability.)
For 0 β€ π₯ β€ πΏ/π search for a solution of the form
π+(π₯) = π(π) π2πππ₯π/πβ²(π) + π(π) πβ2πππ₯π/πβ²(π),
πβ(π₯) = π(π) π2πππ₯π/πβ²(π) + π(π) πβ2πππ₯π/πβ²(π).
for some coefficients π, π, π, π β C and πβ² > 0 depending on π (we have a combination of refracted and reflectedwaves). In what follows π, π, π, π, πβ², π, π‘ denote the limits of the above numbers as π β 0.
The limit πβ² (wavelength of refracted wave) is determined by Diracβs equation as follows. Substituting theexpressions for πΒ±(π₯) into Diracβs equation, canceling common factors, and taking the limit π β 0 we get
ππ =
(πβ 2π
πβ² +2π
π
)π,
ππ =
(π+
2π
πβ² +2π
π
)π.
29
We also get an analogous equation with π and π instead of π and π. This implies that πβ² = π/π, where π =β1 +ππ/π (refractive index ).Sewing together the solutions (namely, π+(π₯) at π₯ = 1 and π₯ = πΏ/π+1, whereas πβ(π₯) at π₯ = 0 and π₯ = πΏ/π)
gives a system of linear equations in π, π, π, π, π, π‘:
πβ(0) = π(π) = π(π) + π(π), πβ(πΏ/π) = 0 = π(π) π2πππΏ/πβ²+ π(π) πβ2πππΏ/πβ²
,
π+(1) = ππ2πππΏ/π = π(π) π2πππΏ/πβ²+ π(π) πβ2πππΏ/πβ²
, π+(πΏ/π+ 1) = π‘(π) π2ππ(πΏ+π)/π = π(π) π2ππ(πΏ+π)/πβ²+ π(π) πβ2ππ(πΏ+π)/πβ²
.
Passing to the limit π β 0 and solving the resulting system of linear equations we find π, π, π, π, π, π‘ Surely, one stillneeds to check that even before passing to the limit, the expressions for πΒ±(π₯) determined by our equations indeedsatisfy Diracβs equation in each black square. We omit that.
It remains to prove that (π+(π₯), πβ(π₯)) = limπ¦β+βπ¦+π₯ even
π 2ππ¦π/ποΏ½οΏ½(π₯,π¦,π3(π₯)π,ππ ,β); then the number |π|2 is the required
limit. We do not know any elementary proof of that assertion and thus omit the details.Informally, the idea is to show that the limit does not depend on the initial values π+(π₯, 0) for π₯ > 0 and
πβ(π₯, 0) for π₯ < πΏ/π; then one can replace these initial values by the solution we have found above and get theresult. Independence on π+(π₯, 0) for π₯ > πΏ/π and πβ(π₯, 0) for π₯ < 0 is obvious. To prove independence onπΒ±(π₯, 0) for 0 β€ π₯ β€ πΏ/π, assume that πΒ±(π₯, 0) vanishes outside the segment 0 β€ π₯ β€ πΏ/π and consider the lineartransformation
π : (πβ(0, 0), πβ(2, 0), . . . , πβ(πΏ/πβ 1, 0), π+(0, 0), . . . , π+(πΏ/πβ 1, 0))
β¦β (πβ(0, 2), πβ(2, 2), . . . , πβ(πΏ/πβ 1, 2), π+(0, 2), . . . , π+(πΏ/πβ 1, 2)).
Let us show that the absolute values of all the eigenvalues of π are less than 1. Consider an eigenvector π£ withthe coordinates π£1 = πβ(0, 0), . . . , π£πΏ/π+1 = π+(πΏ/π β 1, 0). Let π£π be the leftmost nonzero coordinate. If π > 1then ππ£ has nonzero (π β 1)-th coordinate; but ππ£ is proportional to π£, a contradiction. Thus π = 1. Henceπβ(β2, 2) = 0 by Diracβs equation. Then by the probability conservation
βπ£β = |πβ(0, 0)|2 + |πβ(2, 0)|2 + Β· Β· Β·+ |π+(πΏ/πβ 1, 0)|2
= |πβ(β2, 2)|2 + |πβ(0, 2)|2 + Β· Β· Β·+ |π+(πΏ/πβ 1, 2)|2
> |πβ(0, 2)|2 + Β· Β· Β·+ |π+(πΏ/πβ 1, 2)|2
= βππ£β.
Thus the absolute value of the eigenvalue π£ is strictly less than 1. Thus βππ¦β β 0 as π¦ β β, which proves thatthe values πΒ±(π₯, 0) for 0 β€ π₯ β€ πΏ do not affect the limit.
26. Answer : In the table, the vector in the cell (π₯, π¦) is οΏ½οΏ½(π₯, π¦, π’,+), and an empty cell (π₯, π¦) means thatοΏ½οΏ½(π₯, π¦, π’,+) = (0, 0).
4(0, 1
2β2
)(0, 0)
(0, 1
2β2
)3
(0,β1
2
) (0, 12
)2
(0,β 1β
2
)1 (0,β1)
y x β2 β1 0 1 2 3 4
In the following table the number in the cell (π₯, π¦) is π (π₯, π¦, π’,+), and an empty cell (π₯, π¦) means thatπ (π₯, π¦, π’,+) = 0.
4 1/8 0 1/8
3 1/4 1/4
2 1/2
1 1
y x β2 β1 0 1 2 3 4
(The vectors οΏ½οΏ½(π₯, π¦, π’,+) can be easily computed using Problem 31.)For any positive π¦ we have
βπ₯βZ
(π (π₯, π¦, π’,+)+π (π₯, π¦, π’,β)) = 1. (This is one of the statements of Problem 31.)
27. Answer (kindly provided by Gleb Minaev and Ivan Russkikh, participants of Summer Conference of Tourna-ment of Towns):
30
28. Solution. Suppose that π’β² is the result of changing the values of π’ in all the vertices of a black square (π, π).Notice that if a path π neither starts nor ends in the square (π, π), then it either passes through two vertices witha changed value of π’ or does not pass through such vertices at all. Therefore, οΏ½οΏ½(π , π’) = οΏ½οΏ½(π , π’β²). The equationπ (π₯, π¦, π’β²,+) = π (π₯, π¦, π’,+) for any (π₯, π¦) = (π, π) follows immediately.
The only case left to prove is (π, π) = (π₯, π¦) or (π, π) = (0, 0). For each path π starting or ending in (π, π)we have οΏ½οΏ½(π , π’β²) = βοΏ½οΏ½(π , π’) because π passes through exactly one vertex with a changed value of π’. Therefore,π (π, π, π’β²,+) = π (π, π, π’,+).29. Solution. The top-right vertex of the square (π₯, π¦) is called vertex (π₯, π¦). Denote by π’(π₯, π¦) the value of π’ atthe vertex (π₯, π¦). Reversing a black square (π₯, π¦) means changing the values of π’ at all the four vertices of thesquare.
Notice that reversing a black square does not change the sign of any white squares, because the number ofchanged vertices in each white square is even. If all values of a field in a rectangle are positive, then all the whitesquares inside the rectangle are positive. Therefore, it is impossible to make π’ identically +1 by reversing blacksquares, if initially there is a negative white square.
Suppose that there are no negative white squares in a rectangle π Γπ , where π is the height of the rectangle.Let the vertex (0, 0) be at the bottom-left of the rectangle. Let us prove that π’ can be made identically +1 throughreversal of black squares by induction over π .
Base: π = 0. Let us describe the algorithm for making π’ identically +1 in the rectangle π Γ 0 (i.e in vertices(0, 0), (1, 0), . . . , (π, 0)).
Note that for all π we can reverse either square (π+ 1, 0) or square (π+ 1, 1), thereby we can change the signsof the values of π’ in vertices (π, 0) and (π + 1, 0). Denote this operation reversing π.
The algorithm is as follows: Find the minimum 0 β€ π β€ π such that π’(π, 0) = β1, and reverse π. If after thatthere still exists 0 β€ π β€ π such that π’(π, 0) = β1, than the minimum such π is at least 1 more than before usingthe operation. Therefore, after at most π + 1 operations we make π’ to be identically +1 in the rectangle π Γ 0.
Step: Suppose that the statement is true for π and prove it for π + 1 by giving an explicit algorithm.First make π’ identically +1 in the lower rectangle π Γ π by applying the algorithm for π . Then for each
white square (π,π + 1) in the rectangle reverse the square (π,π + 2) if π’(π β 1, π + 1) = π’(π,π + 1) = β1,otherwise do nothing. Finally, if the square (0, π + 1) is black and π’(β1, π + 1) = β1 then reverse the square(β1, π +2), and if the square (π,π +1) is black and π’(π,π +1) = β1 then reverse the square (π +1, π +2).The result is a field, positive in the whole rectangle (π + 1)Γπ .30. Answer: Zero, if one of the sides of the rectangle is 1, and any even number no greater than the number ofwhite squares in the rectangle, if both sides are greater than 1.
Hint. Denote by π(π₯, π¦, π’) the number of vertices of the square (π₯, π¦) where the value of π’ is negative. Noticethat any vertex that is strictly inside the rectangle (not on the boundary) is a vertex of exactly four squares inthat rectangle, two of which are white. Therefore, the sum of π(π₯, π¦, π’) over all white squares in the rectangle iseven. Hence the number of negative white squares is even.
Let us show that any even number no greater than the number of white squares in the rectangle is indeedpossible.
If both the length and the width of the rectangle are even, then for any field that is positive everywhere in therectangle except for π vertices with both coordinates even there are exactly 2π negative white squares.
For odd length or width (greater than 1) see the figure below. The identically +1 field has no negative whitesquares. The field π’2π+2 with 2π + 2 negative white squares is obtained from the field π’2π by changing all thevalues of π’2π to the opposite in any single encircled area where they have not been changed before.
31
31. Answer. Recall that π’(π₯, π¦) is the value of π’ in the top-right vertex of the square (π₯, π¦). Then
π1(π₯, π¦, π’) =1β2π’(π₯, π¦ β 1)(π1(π₯+ 1, π¦ β 1, π’) + π2(π₯+ 1, π¦ β 1, π’)),
π2(π₯, π¦, π’) =1β2π’(π₯β 1, π¦ β 1)(βπ1(π₯β 1, π¦ β 1, π’) + π2(π₯β 1, π¦ β 1, π’)).
Hint. The solution of the analogue of Problem 4 is very similar to the original one, only the factors π’(π₯, π¦β 1)and π’(π₯ β 1, π¦ β 1) are added due to the last step of the path π passing through the top-right vertex of either(π₯, π¦ β 1) or (π₯β 1, π¦ β 1).
Let us prove the analogue of Problem 5 by induction over π¦. The step of induction follows immediately fromthe following computation:β
π₯βZπ (π₯, π¦ + 1, π’) =
βπ₯βZ
[π1(π₯, π¦ + 1, π’)2 + π2(π₯, π¦ + 1, π’)2
]=
βπ₯βZ
π1(π₯, π¦ + 1, π’)2 +βπ₯βZ
π2(π₯, π¦ + 1, π’)2 =
=1
2
βπ₯βZ
π’(π₯, π¦)2(π1(π₯+ 1, π¦, π’) + π2(π₯+ 1, π¦, π’))2 +1
2
βπ₯βZ
π’(π₯β 1, π¦)2(π2(π₯β 1, π¦, π’)β π1(π₯β 1, π¦, π’))2 =
=βπ₯βZ
(π1(π₯, π¦, π’) + π2(π₯, π¦, π’))2
2+βπ₯βZ
(π2(π₯, π¦, π’)β π1(π₯, π¦, π’))2
2=
βπ₯βZ
[π1(π₯, π¦, π’)
2 + π2(π₯, π¦, π’)2]=
βπ₯βZ
π (π₯, π¦, π’).
32. Answer: π (π΄π΅,π΄β²π΅β² β πΈπΉ,πΈβ²πΉ β²) =
β§βͺβͺβͺβͺβ¨βͺβͺβͺβͺβ©π (π₯, π¦,β)π (π₯β² β π₯0, π¦,β), if πΈ = (π₯+ 1, π¦), πΈβ² = (π₯β² + 1, π¦),
π (π₯, π¦,β)π (π₯β² β π₯0, π¦,+), if πΈ = (π₯+ 1, π¦), πΈβ² = (π₯β² β 1, π¦),
π (π₯, π¦,+)π (π₯β² β π₯0, π¦,β), if πΈ = (π₯β 1, π¦), πΈβ² = (π₯β² + 1, π¦),
π (π₯, π¦,+)π (π₯β² β π₯0, π¦,+), if πΈ = (π₯β 1, π¦), πΈβ² = (π₯β² β 1, π¦).
Solution. Notice that οΏ½οΏ½(π , π β²) = βππ(π )π(π β²), where οΏ½οΏ½π denotes the product of complex numbers οΏ½οΏ½ and οΏ½οΏ½.Without loss of generality suppose that π₯β² > π₯. Due to the condition π₯0 β₯ 2π¦ there are no paths starting at π΄ andending in πΉ β² and no paths starting at π΄β² and ending in πΉ .
Thereforeβ
π :π΄π΅βπΈβ²πΉ β²π β²:π΄β²π΅β²βπΈπΉ
οΏ½οΏ½(π , π β²) = 0 and οΏ½οΏ½(π΄π΅,π΄β²π΅β² β πΈπΉ,πΈβ²πΉ β²) =β
π :π΄π΅βπΈπΉπ β²:π΄β²π΅β²βπΈβ²πΉ β²
οΏ½οΏ½(π , π β²).
If πΈ = (π₯β 1, π¦) and πΈβ² = (π₯β² β 1, π¦) then οΏ½οΏ½(π΄π΅,π΄β²π΅β² β πΈπΉ,πΈβ²πΉ β²) = βπβπ ,π β²
οΏ½οΏ½(π )π(π β²), where the sum is over
π and π β² ending with an upwards-right move. Therefore, οΏ½οΏ½(π΄π΅,π΄β²π΅β² β πΈπΉ,πΈβ²πΉ β²) = βππ(π₯, π¦,+)π(π₯β² β π₯0, π¦,+).Hence π (π΄π΅,π΄β²π΅β² β πΈπΉ,πΈβ²πΉ β²) = π (π₯, π¦,+)π (π₯β² β π₯0, π¦,+).
If πΈ = (π₯+1, π¦) and πΈβ² = (π₯β²β 1, π¦) then οΏ½οΏ½(π΄π΅,π΄β²π΅β² β πΈπΉ,πΈβ²πΉ β²) = βπβπ ,π β²
οΏ½οΏ½(π )π(π β²), where the sum is over π
ending with an upwards-left move and π β² ending with an upwards right move. Hence π (π΄π΅,π΄β²π΅β² β πΈπΉ,πΈβ²πΉ β²) =π (π₯, π¦,β)π (π₯β² β π₯0, π¦,+).
If πΈ = (π₯β 1, π¦) and πΈβ² = (π₯β²+1, π¦) then οΏ½οΏ½(π΄π΅,π΄β²π΅β² β πΈπΉ,πΈβ²πΉ β²) = βπβπ ,π β²
οΏ½οΏ½(π )π(π β²), where the sum is over π
ending with an upwards-right move and π β² ending with an upwards left move. Hence π (π΄π΅,π΄β²π΅β² β πΈπΉ,πΈβ²πΉ β²) =π (π₯, π¦,+)π (π₯β² β π₯0, π¦,β).
32
If πΈ = (π₯+ 1, π¦) and πΈβ² = (π₯β² + 1, π¦) then οΏ½οΏ½(π΄π΅,π΄β²π΅β² β πΈπΉ,πΈβ²πΉ β²) = βπβπ ,π β²
οΏ½οΏ½(π )π(π β²), where the sum is over
π and π β² ending with an upwards-left move. Hence π (π΄π΅,π΄β²π΅β² β πΈπΉ,πΈβ²πΉ β²) = π (π₯, π¦,β)π (π₯β² β π₯0, π¦,β).
33. Hint. Suppose that paths π : π΄π΅ β πΈπΉ and π β² : π΄β²π΅β² β πΈβ²πΉ β² have a common move. Suppose that πΆπ· is thefirst common move of π and π β². Denote by π the path that coincides with π up to the square πΆ and with π β² after πΆ.and π β² the path that coincides with π β² before πΆ and with π after. The paths π , π β² start with the moves π΄π΅,π΄β²π΅β²
and end with moves πΈβ²πΉ β² and πΈπΉ respectively.The total number of turns in the paths π and π β² equals the total number of turns in π and π β² , because πΆπ· is a
common move for all of them. Therefore, οΏ½οΏ½(π , π β²) = οΏ½οΏ½(π , π β²). Notice that π : (π , π β²) β¦β (π , π β²) is a bijection, becauseπΆπ· is exactly the first common move. One of (π , π β²) and (π , π β²) contributes to οΏ½οΏ½(πΈπΉ,πΈβ²πΉ β²) with the coefficient(+1) and the other one with the the coefficient (β1). Therefore, these pairs cancel each other.
34. Hint. Let us prove this equality by induction over π¦. The base is obvious.Note that by analogy with Problem 11, οΏ½οΏ½(π΄π΅,π΄β²π΅β² β πΈπΉ,πΈβ²πΉ β²) is parallel to either the line π₯ = 0 or the line
π¦ = 0. Denote by π1(π΄π΅,π΄β²π΅β² β πΈπΉ,πΈβ²πΉ β²) and π2(π΄π΅,π΄β²π΅β² β πΈπΉ,πΈβ²πΉ β²) its coordinates.Fix the starting moves π΄π΅ and π΄β²π΅β² of the paths. Denote by π(πΈ,πΈβ², β,+,β) = π1(π΄π΅,π΄β²π΅β² β πΈπΉ,πΈβ²πΉ β²),
where πΈπΉ is an upwards-right move, πΈβ²πΉ β² is an upwards-left move, and π1(π΄π΅,π΄β²π΅β² β πΈπΉ,πΈβ²πΉ β²) is the (possibly)non-zero coordinate of the corresponding vector. π(πΈ,πΈβ², β,+,+), π(πΈ,πΈβ², β,β,+) and π(πΈ,πΈβ², β,β,β) are definedanalogously. Denote π(πΈ,πΈβ², β,+,β) = π1(π΄π΅,π΄β²π΅β² β π·πΈ,π·β²πΈβ²), where π·πΈ is an upwards-right move and π·β²πΈβ²
is an upwards-left move. π(πΈ,πΈβ², β,+,+), π(πΈ,πΈβ², β,β,+) and π(πΈ,πΈβ², β,β,β) are defined analogously.Note that the induction step is a proof of the equalityβ
πΈ,πΈβ²
((π(πΈ,πΈβ², β,+,+))2 + (π(πΈ,πΈβ², β,+,β))2 + (π(πΈ,πΈβ², β,β,+))2 + (π(πΈ,πΈβ², β,β,β))2) =
=βπΈ,πΈβ²
((π(πΈ,πΈβ², β,+,+))2 + (π(πΈ,πΈβ², β,+,β))2 + (π(πΈ,πΈβ², β,β,+))2 + (π(πΈ,πΈβ², β,β,β))2).
We will prove this equation by proving that for every pair (πΈ,πΈβ²), where πΈ and πΈβ² have the same π¦-coordinate,
(π(πΈ,πΈβ², β,+,+))2 + (π(πΈ,πΈβ², β,+,β))2 + (π(πΈ,πΈβ², β,β,+))2 + (π(πΈ,πΈβ², β,β,β))2 =
=(π(πΈ,πΈβ², β,+,+))2 + (π(πΈ,πΈβ², β,+,β))2 + (π(πΈ,πΈβ², β,β,+))2 + (π(πΈ,πΈβ², β,β,β))2.
Analogously to Problem 4 we obtain
π(πΈ,πΈβ², β,+,+) =1
2(π(πΈ,πΈβ², β,+,+)β π(πΈ,πΈβ², β,+,β)β π(πΈ,πΈβ², β,β,+)β π(πΈ,πΈβ², β,β,β)),
π(πΈ,πΈβ², β,+,β) =1
2(π(πΈ,πΈβ², β,+,+)β π(πΈ,πΈβ², β,+,β) + π(πΈ,πΈβ², β,β,+) + π(πΈ,πΈβ², β,β,β)),
π(πΈ,πΈβ², β,β,+) =1
2(π(πΈ,πΈβ², β,+,+) + π(πΈ,πΈβ², β,+,β)β π(πΈ,πΈβ², β,β,+) + π(πΈ,πΈβ², β,β,β)),
π(πΈ,πΈβ², β,β,β) =1
2(βπ(πΈ,πΈβ², β,+,+)β π(πΈ,πΈβ², β,+,β)β π(πΈ,πΈβ², β,β,+) + π(πΈ,πΈβ², β,β,β)).
With this, we can prove our equation by substitution and expansion.
35. Answer: π (π΄π΅,π΅β²π΄β² β πΈπΉ,πΉ β²πΈβ²) =
β§βͺβͺβͺβͺβ¨βͺβͺβͺβͺβ©π (π₯, π¦,β)π (π₯0 β π₯β², π¦,β), if πΈ = (π₯+ 1, π¦), πΈβ² = (π₯β² + 1, π¦),
π (π₯, π¦,β)π (π₯0 β π₯β², π¦,+), if πΈ = (π₯+ 1, π¦), πΈβ² = (π₯β² β 1, π¦),
π (π₯, π¦,+)π (π₯0 β π₯β², π¦,β), if πΈ = (π₯β 1, π¦), πΈβ² = (π₯β² + 1, π¦),
π (π₯, π¦,+)π (π₯0 β π₯β², π¦,+), if πΈ = (π₯β 1, π¦), πΈβ² = (π₯β² β 1, π¦).
Hint. Notice that οΏ½οΏ½(π , π β²) = ππ(π )π*(π β²), where οΏ½οΏ½π denotes the product of complex numbers οΏ½οΏ½ and οΏ½οΏ½, and οΏ½οΏ½*
denotes conjugation.For the remaining problems, we only give preliminary draft solutions.
36. Answer: for each π’fin we have
π (2, 2, π’fin, 1,+) = 1/4, π (0, 2, π’fin, 1,β) = 1/4,
π (3, 3, π’fin, 1,+) = 1/32, π (1, 3, π’fin, 1,β) = 1/32,
π (1, 3, π’fin, 1,+) = 1/32, π (β1, 3, π’fin, 1,β) = 1/32;
all the other probabilities vanish for π¦ = 2, 3.
37. Answer :βπ’οΏ½οΏ½(π, π , π’) = 23π¦β4(1 + π2)
π¦β12 cos((π β π) arctan(π))π(π ).
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Hint. For brevity, denote π = 1
2(π¦β2)2 (1+π2)(π¦β1)2
2
.
Suppose that π and π are the numbers of white squares in π to the right and to the left from π respectively.Notice that π and π are odd if and only if the number of βββ signs on the path π is odd (that is, οΏ½οΏ½(π , π’) = βοΏ½οΏ½(π )).
Easy-to-prove lemma: if π is the number of black squares in π not touching the boundary, then the numberof fields π’ with exactly 2π negative white squares equals 2π times the number of ways to assign βββ to 2π whitesquares in π . The factor 2π is the number of gauge transformations from the Problem 28.
By the lemma we have
βπ’
οΏ½οΏ½(π, π , π’) = πβπ’
(βππ)ποΏ½οΏ½(π , π’) = 2ππ
β‘β£ βπ,π ππ£ππ
(βππ)π+π
(π
π
)(π
π
)β
βπ,π πππ
(βππ)π+π
(π
π
)(π
π
)β€β¦ οΏ½οΏ½(π ) =
= 2πβ2π[((1 + ππ)π + (1β ππ)π)
((1 + ππ)π + (1β ππ)π
)β ((1 + ππ)π β (1β ππ)π)
((1 + ππ)π β (1β ππ)π
)]οΏ½οΏ½(π ) =
= 2πππ π((1 + ππ)π(1β ππ)π
)οΏ½οΏ½(π ) = 2π(1 + π2)
π+π2 π cos((π β π) arctan(π))π(π ).
Since π = π + π = π¦2 β π¦, it follows thatβπ’οΏ½οΏ½(π, π , π’) = 23π¦β4(1 + π2)
π¦β12 cos((π β π) arctan(π))π(π ).
38. Answer. Denote
πββ(π₯, π¦) = π2(π₯, π¦, π’β, π,β), πβ+(π₯, π¦) = π1(π₯, π¦, π’β, π,+),
π+β(π₯, π¦) = π1(π₯, π¦, π’+, π,β), π++(π₯, π¦) = π2(π₯, π¦, π’+, π,+).
Denote by π (π¦) = π the rectangle formed by all the squares (π₯β², π¦β²) such that 1 β π¦ < π₯β² < π¦ and 0 < π¦β² < π¦.Denote πΌ = πΌ(π₯, π¦) = (Ξπ β Ξπ) arctan π, where Ξπ and Ξπ are the numbers of white squares in π (π¦) outsideπ (π¦ β 1) lying to the left and to the right from the square (π₯+ 1, π¦ β 1). Denote π½ = πΌ(π₯β 2, π¦). Thenβ§βͺβͺβͺβͺβͺβ¨βͺβͺβͺβͺβͺβ©
π++(π₯, π¦) = cosπ½β2(π++(π₯β 1, π¦ β 1)β π+β(π₯β 1, π¦ β 1))β sinπ½β
2(πβ+(π₯β 1, π¦ β 1)β πββ(π₯β 1, π¦ β 1)) ;
π+β(π₯, π¦) = cosπΌβ2(π++(π₯+ 1, π¦ β 1) + π+β(π₯+ 1, π¦ β 1))β sinπΌβ
2(πβ+(π₯+ 1, π¦ β 1) + πββ(π₯+ 1, π¦ β 1)) ;
πβ+(π₯, π¦) = sinπ½β2(π++(π₯β 1, π¦ β 1)β π+β(π₯β 1, π¦ β 1)) + cosπ½β
2(πβ+(π₯β 1, π¦ β 1)β πββ(π₯β 1, π¦ β 1)) ;
πββ(π₯, π¦) = sinπΌβ2(π++(π₯+ 1, π¦ β 1) + π+β(π₯+ 1, π¦ β 1)) + cosπΌβ
2(πβ+(π₯+ 1, π¦ β 1) + πββ(π₯+ 1, π¦ β 1)) .
Hint. The latter formula defines an orthogonal transformation, which implies thatβ
π₯βZ(π (π₯, π¦, π,+) +π (π₯, π¦, π,β)) = 1 by induction over π¦.
39. Answer. The number assigned to a basis configuration is shown below it:
1 βπππβ1+π2π2
1β1+π2π2
1β1+π2π2
βπππβ1+π2π2
β 1.
Hint. Let us prove that the sum over all pairs of paths passing through vertices with βββ signs only equals theproduct of all the numbers assigned to the black squares in the rectangle.
Notice that the squares of type (1) do not contribute to the result at all. Squares of types (2) and (5) canonly be passed through by a path with a turn in that square, which means contributing a factor of βπππβ
1+π2π2to
οΏ½οΏ½(π , π β²). Squares of types (3) and (4) can only be passed through by a path with a turn in that square which meanscontributing a factor of 1β
1+π2π2to οΏ½οΏ½(π , π β²). Squares of type (6) can be passed through in two different ways: by
two turning or two not turning paths. When the paths do turn, the square contributes(
βπππβ1+π2π2
)2and when they
do not, it contributes β11+π2π2
, which sums up to β1.
Acknowledgements
The authors are grateful to V. Skopenkova for some of the figures, G. Chelnokov, I. Ivanov for useful discus-sions, and all participants of Summer Conference of Tournament of Towns in Arandelovac and Summer School inContemporary Mathematics in Dubna for their contribution.
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